IN  MEMORIAM 
FLOR1AN  CAJOR1 


PLANE  AND   SPHERICAL  TRIGONOMETRY 


THE  MACMILLAN  COMPANY 

NEW  YORK   •    BOSTON    •    CHICAGO   •    DALLAS 
ATLANTA  •    SAN   FRANCISCO 

MACMILLAN  &  CO.,  LIMITED 

LONDON  •    BOMBAY  •    CALCUTTA 
MELBOURNE 

THE  MACMILLAN  CO.  OF  CANADA,  LTD. 

TORONTO 


PLANE   AND    SPHERICAL 
TRIGONOMETRY 


BY 


LEONARD   M.    PASSANO 

ASSOCIATE    PROFESSOR    OF    MATHEMATICS 
MASSACHUSETTS    INSTITUTE    OF    TECHNOLOGY 


Nefo  fgotfc 

THE   MACMILLAN   COMPANY 
1918 

All  rights  reserved 


COPYRIGHT,  1918, 
BY  THE  MACMILLAN  COMPANY. 


Set  up  and  electrotypcd.    Published  April,  1918. 


NorfaooU 

J.  S.  Gushing  Co.  —  Berwick  &  Smith  Co. 
Norwood,  Mass.,  U.S.A. 


•    PREFACE 

OF  late  years,  in  the  writing  of  textbooks  of  trigonometry, 
a  tendency  to  amplification  has  shown  itself,  doubtless  with 
the  idea  that  amplification  means  simplification.  Unfortu- 
nately the  amplification  has  spent  itself  upon  details  rather 
than  upon  principles,  which  latter  have  too  often  been  in- 
adequately treated.  The  result  has  been  textbooks  which 
overlook  the  comparative  maturity  of  the  boys  and  girls 
who  study  trigonometry  and  which  cling  almost  with  affec- 
tion to  the  practices  of  the  most  elementary  mathematics. 

The  present  text'  aims  to  present  the  trigonometry  in 
such  a  way  as  to  make  it  interesting  to  students  approach- 
ing some  maturity,  and  so  as  to  connect  the  subject,  not 
only  with  the  mathematics  which  the  student  has  already 
had,  but  also  with  the  mathematics  which,  in  many  cases  at 
least,  is  to  follow.  A  subject  may  be  so  burdened  with 
detailed  explanations  as  to  become  monotonous  and  lifeless, 
or,  on  the  other  hand,  presented  in  so  concise  and  difficult  a 
manner  as  to  be  repellent.  The  present  work  endeavors  to 
avoid  both  extremes.  Full  explanations  are  given  of  im- 
portant principles,  but  many  simple  details  are  left  to  the 
work  of  the  student. 

The  following  points  in  the  text  may  be  noted : 

1.  Positive  and  negative  angles  of  any  magnitude  and  the 
trigonometric  functions  of  such  angles,  defined  by  means  of 
a  system  of  rectangular  coordinates,  are  taken  up  in  the 
beginning  of  the  book ;  acute  angles,  with  their  functions, 
being  mentioned  as  a  special  case. 

2.  Thus  the  basic  trigonometric  identities  are  got  at  once 
for  all  angles. 


vi  PREFACE 

3.  The  functions  of  0°,  90°,  etc.,  are  carefully  explained 
by  the  theory  of  limits. 

4.  The  solution  of  right  triangles  and  related  problems 
are  taken  up  early  without  the  use  of  logarithms. 

5.  Logarithms  are  then  very  carefully  explained  and  fully 
discussed,  not  so  much  as  to  their  use  in  computation,  but 
rather  so  as  to  clarify  their  meaning. 

6.  Right  triangles  are  then  solved  by  the  use  of  loga- 
rithms, and  the  essentially  approximate  nature  of  all  nu- 
merical results  is  emphasized. 

7.  The  text  next  returns  to  trigonometric  identities,  giv- 
ing a  detailed  and  accurate  proof  of  the  addition  formulae 
for  sines  and  cosines,  with  less  detailed  but  sufficient  expla- 
nation of  other  fundamental  identities.      The  number  of 
identities  to  be  memorized  is  reduced  to  a  minimum. 

8.  The  circular  measure  of  an  angle  and  the  inverse  func- 
tions are  then  taken  up,  emphasis  being  laid  upon  the  fact 
that  the  latter  are  angles. 

9.  There  follows  the  solution  of  triangles  in  general.     As 
each  case  is  mentioned  the  theorems  or  formulas  needed  for 
its  solution  are  derived. 

10.  The  last  subject  treated  in  the  plane  trigonometry  is 
the  solution  of  trigonometric  equations,  and  the  fact  is  em- 
phasized  that  the  operations  are  simply  the    solution   of 
algebraic  equations  applied  to  a  new  class  of  quantities. 

11.  The  lists  of  examples  and  problems  are  numerous 
and  carefully  chosen,  many  of  them  being  taken  from  work 
in  analytic  geometry  and  calculus,  though,  of  course,  no 
knowledge  of  either  of  these  subjects  is  assumed.     Some  of 
the  problems  are  entirely  new,  being  invented  for  this  text, 
and  all  problems  are  chosen  with  a  purpose  to  indicate  the 
practical  interest  and  value  of  trigonometry. 

12.  In  the  spherical  trigonometry,  as  in  the  plane,  the 
three   chief  aims   are   brevity,   clarity,  and  simplicity ;    a 
chapter  on  the  Earth  treated  as  a  sphere  being  given  to 
enliven  an  otherwise  somewhat  formal  and  lifeless  subject. 

13.  The  author  has  not  tried  to  revolutionize  the  teaching 


PREFACE  vii 

of  trigonometry,  believing  that  much  that  has  been  done  in 
the  past  is  good  though  none  the  less  open  to  improvement. 
Such  improvement  has  been  the  aim  of  this  work. 

The  author  wishes  to  acknowledge  the  kindness  of  his 
colleagues  Professor  H.  W.  Tyler,  Professor  F.  L.  Hitch- 
cock, and  Professor  J.  Lipka  in  reading  and  criticizing  the 
manuscript  of  his  book,  and  to  express  his  thanks  to  Pro- 
fessor E.  R.  Hedrick,  editor  of  the  tables  appended,  for 
permission  to  make  use  of  them. 

L.  M.  PASSANO. 


CONTENTS 

XBT.  PAGE 

INTRODUCTION  .........  xiii 

PLANE  TRIGONOMETRY 

CHAPTER  I.     THE  TRIGONOMETRIC  FUNCTIONS  OF  ANY 

ANGLE  AND  IDENTICAL  RELATIONS  AMONG  THEM    .  1-13 

1.  Rectangular  coordinates 1 

2.  Angles  of  any  magnitude 1 

3.  Abscissa,  ordinate,  and  distance 3 

4-5.   The  trigonometric  functions  denned     ....  4 

6.  Signs  of  the  functions 6 

7.  Functions  of  acute  angles         .......  7 

8.  Reciprocal  functions 7 

9.  Tangent,  sine,  and  cosine 8 

10.  Sine  and  cosine         ........  9 

11.  Tangent  and  secant ........  9 

12-13.   Fundamental  relations,  collected        ....  10 

14.  Values  of  the  functions  when  one  is  given       ...  10 

CHAPTER  II.     IDENTICAL  RELATIONS  AMONG  THE  FUNC- 
TIONS OF  RELATED  ANGLES.     THE  VALUES  OF  THE 

FUNCTIONS  OF  CERTAIN  ANGLES      ....  14-27 

15.  Functions  of  negative  angles 14 

16.  Functions  of  90°  —  a 15 

17.  Functions  of  90°  +  a        .        .        .        .        .        .        .  16 

18.  Functions  of  180°  +  a 17 

19.  Generalization 18 

20.  Functions  of  certain  angles 20 

21.  Functions  of  30°  and  60° 20 

22.  Functions  of  45° 21 

23.  Functions  of  120°,  135°,  etc 22 

24.  Functions  of  0° 22 

25-26.    Functions  of  90°,  etc. 25 

27.   Limiting  values  of  functions 26 


X  CONTENTS 

ART.  PAGE 

CHAPTER  III.  THE  SOLUTION  OF  EIGHT  TRIANGLES. 
LOGARITHMS  AND  COMPUTATION  BY  MEANS  OF 
LOGARITHMS 28-44 

28.  Solution  of  right  triangles 28 

29.  Logarithms 30 

20.   The  common  system 31 

31.  Mantissa  and  characteristic 32 

32.  Four  computation  theorems 34 

33.  Special  properties  of  logarithms 36 

34-35.   Illustrative  examples 37 

CHAPTER  IV.     FUNDAMENTAL  IDENTITIES       .         45-58 

36.    Introductory 45 

37-40.   The  addition  formulae,  sine  and  cosine       .         .         .  45 

41.   The  addition  formulae,  tangent  and  cotangent         .        .  50 

42-43.   Functions  of  the  double  angle     .....  51 

44.  Functions  of  the  half  angle      .......  52 

45.  Sum  of  sines  ;  sum  of  cosines  ......  53 

46.  Identities  and  equations  .......  64 

CHAPTER  V.     THE  CIRCULAR  OR  RADIAN  MEASURE  OF 

AN  ANGLE.     INVERSE  TRIGONOMETRIC  FUNCTIONS     .         59-67 

47.  Circular  or  radian  measure      ......  59 

48.  Inverse  functions      ........  62 

49.  General  value  of  an  angle         .        .        .         .        .        .  63 

CHAPTER  VI.     THE  SOLUTION  OF  GENERAL  TRIANGLES         68-83 

50.  Four  cases 68 

51.  The  law  of  sines 68 

52.  Case  I ;  a  side  and  two  angles          .....  69 

53.  Case  II ;  two  sides  and  an  opposite  angle        ...  70 
54-55.    Case  III ;  two  sides  and  included  angle      ...  71 

56.   The  law  of  cosines 73 

57-59.   Formulae  from  the  law  of  cosines        ....  74 

66.    Case  IV ;  three  sides 76 

61.  Case  III ;  other  methods  of  solution        ....  77 

62.  Areas  of  right  triangles 78 

63.  Areas  of  oblique  triangles 79 

CHAPTER    VII.      THE     SOLUTION    OF    TRIGONOMETRIC 

EQUATIONS 84-93 

64.  Introductory 84 

65.  Illustrative  examples 84 


CONTENTS  xi 

ART.  PAGE 

66.  Special  types  of  equations 86 

67.  Simultaneous  equations 89 

68.  Inverse  trigonometric  equations 90 


SPHERICAL  TRIGONOMETRY 

CHAPTER  VIII.     FUNDAMENTAL  RELATIONS     .  94-99 

69.  Introductory 94 

70.  The  law  of  cosines 94 

71.  The  law  of  cosines  ;  polar  triangle 97 

72.  The  law  of  sines 98 

CHAPTER   IX.      THE    SOLUTION   OF   RIGHT   SPHERICAL 

TRIANGLES 100-106 

73.  Special  formulae  for  right  triangles  .....  100 

74.  Special  formulae ;  collected.     Napier's  rules   .        .        .  101 

75.  Rules  for  solution 102 

76.  Illustrative  examples.     Quadrantal  triangles  .        .        .  103 

CHAPTER  X.     THE   SOLUTION  OF   OBLIQUE   SPHERICAL 

TRIANGLES 107-124 

77.  Six  cases 107 

78.  Case  1 .    Three  sides 107 

79.  Case  2.     Three  angles 110 

80.  Case  3.     Two    sides,    included   angle.     Case  4.     Two 

angles,  included  side.     Napier's  analogies        .         .  Ill 

81.  Illustrative  examples 114 

82.  Case  5.     Two  sides,  opposite   angle.     Case  6.     Two 

angles,  opposite  side 116 

83.  Delambre's  or  Gauss's  equations 120 

CHAPTER  XI.     THE  EARTH  AS  A  SPHERE    .         .     125-131 

84.  Distances  on  the  earth 125 

85.  Position  and  direction  ;  latitude  and  longitude;    course  125 
86-87.   Bearings ;  illustrative  examples          ....  126 
88.   Area  of  a  triangle 129 

ANSWERS 133 


INTRODUCTION 

TRIGONOMETRY  is  primarily  the  science  concerned  with 
the  measurement  of  plane  and  spherical  triangles,  that  is, 
with  the  determination  of  three  of  the  parts  of  such  tri- 
angles when  the  numerical  values  of  the  other  three  parts 
are  given.  This  is  done  by  means  of  the  six  trigono- 
metric functions,  denned  in  article  4  following.  But  these 
functions  enter  so  intimately  into  many  branches  of  mathe- 
matical and  physical  science  not  directly  concerned  with  the 
measurement  of  angles,  that  their  analytical  properties  are 
of  fundamental  importance.  Analytical  trigonometry,  that 
is,  the  proof  and  use  of  various  algebraic  relations  among 
the  trigonometric  functions  of  the  same  or  related  angles, 
is  therefore,  in  modern  times,  of  equal  importance  with  the 
trigonometry  which  deals  with  triangular  solutions. 

The  same  functions  which  enable  one  to  solve  triangles 
constructed  in  a  plane  suffice  also  for  the  solution  of  spheri- 
cal triangles.  But  the  solution  of  triangles  of  which  the 
sides  are  geodetic  lines,  that  is,  lines  which  are  the  shortest 
distances  between  pairs  of  points  on  the  surface,  on  a  sphe- 
roidal surface  such  as  the  Earth,  requires  the  use  of  other 
functions  than  those  needed  for  the  solution  of  plane  or 
spherical  triangles.  This  spheroidal  trigonometry  is  very 
complex,  and  becomes  necessary  only  in  the  accurate  sur- 
vey of  very  large  tracts  of  the  Earth's  surface.  For  ordi- 
nary purposes  of  surveying  and  for  the  solution  of  triangles 
on  the  Earth's  surface  over  small  areas,  plane  and  spherical 
trigonometry  are  sufficient. 

The  study  of  trigonometry,  as  ancillary  to  astronomy, 
dates  from  very  early  times.  Among  the  Greeks,  who, 

xiii 


xiv  INTRODUCTION 

however,  were  more  famous  as  geometers  than  as  investi- 
gators in  other  branches  of  mathematics,  the  names  of 
Hipparchus  (about  150  B.C.)  and  of  Ptolemy  (who  lived  in 
the  second]  century  of  the  Christian  era),  both  astronomers, 
are  prominent.  Hipparchus  left  no  mathematical  writings, 
but  we  are  told  by  an  ancient  writer  that  he  created  the 
science  of  trigonometry.  Ptolemy,  making  use  of  the  inves- 
tigations and  discoveries  of  Hipparchus,  perfected  the  form 
of  the  science.  The  theorems  of  these  two  astronomers  are 
still  the  basis  of  trigonometry. 

Ptolemy  calculated  a  table  of  chords,  which  were  used  in 
those  earliest  days  of  the  science,  as  we  now  use  the  sines 
of  angles.  The  radius  of  a  circle  he  divided  into  sixty 
equal  parts.  Each  of  these  he  divided  again  into  sixty  equal 
parts,  called,  in  the  Latin  translation  of  his-  work  the 
Almagest,  "  partes  minutae  primae " ;  and  each  of  these  in 
turn  into  sixty,  called  "  partes  minutae  secundae " ;  whence 
have  come  the  names  "  minutes "  and  "  seconds "  for  the 
subdivisions  of  the  angular  degree.  Ptolemy,  however,  was 
not  the  first  to  calculate  a  table  of  chords,  Hipparchus, 
among  others,  having  done  so  previously,  but  he  invented 
theorems  by  means  of  which  the  calculations  could  be  more 
readily  made. 

The  Hindus,  more  skillful  calculators  than  the  Greeks, 
acquired  the  knowledge  of  the  latter  and  improved  upon  it, 
notably  in  that  they  calculated  tables  of  the  half-chord,  or 
sine,  instead  of  the  whole  chord  of  the  angle.  The  Arabs 
also  were  acquainted  with  the  Almagest,  and  with  the 
investigations  of  the  Hindus.  It  was  an  Arab,  Al  Battani 
or  Albategnius,  who  first  calculated  a  table  of  what  may  be 
called  cotangents,  by  computing  the  lengths  of  shadows  of 
a  vertical  object  cast  by  the  sun  at  different  altitudes. 
Another  Arab  invented,  as  a  separate  function,  the  tangent, 
which  had  previously  been  used  only  as  an  abbreviation  of 
the  ratio  sine  to  cosine.  Curiously  enough  this  invention 
was  afterwards  forgotten  until  the  tangent  was  re-invented 
in  England  in  the  fourteenth  century  by  Bradwardine,  and 


< 


INTRODUCTION  XV 

in  the  fifteenth  century  by  the  German,  Johannes  Mtiller, 
called  Regiomontanus,  who  wrote  the  first  complete  Euro- 
pean treatise  on  trigonometry. 

When  Napier  *  invented  logarithms,  in  1614,  they  were  at 
once  adopted  in  trigonometric  calculations,  and  the  first 
tables  of  logarithmic  sines  and  tangents  were  made  by 
Edmund  Gunter,  an  English  astronomer  (1581-1626).  He 
it  was  who  first  used  the  names  cosine,  cotangent,  and  co- 
secant. During  the  following  century  the  science  of  trigo- 
nometry progressed  slowly,  becoming  more  analytical  in 
form,  until,  in  the  hands  of  Euler  (1707-1783),  it  became 
essentially  what  it  is  at  the  present  day. 

With  this  brief  introduction  to  the  history  of  trigonom- 
etry let  us  now  proceed  to  become  acquainted  with  that 
homely,  perhaps,  but  most  serviceable  handmaid  to  so  many 
of  the  arts  and  sciences, 

"...  being  just  as  great,  no  doubt, 
Useful  to  men,  and  dear  to  God,  as  they  I  " 

*  John  Napier,  1550-1617. 


'. ' 


PLANE  TRIGONOMETRY 


CHAPTER   I 

THE  TRIGONOMETRIC  FUNCTIONS  OF  ANY  ANGLE,  AND 
IDENTICAL  RELATIONS  AMONG  THEM 

1.  Rectangular  Coordinates.  Two  lines,  x'x  and  y'y,  drawn 
in  a  plane  at  right  angles  to  each  other,  as  in  Fig.  1,  form 
a  system  of  rectangular,  Cartesian  coordinates.  The  point 
0  in  which  the  lines  intersect  is  called  the  origin ;  the  two 


lines  are  called  the  axes  of  coordinates.  One  of  these, 
usually  the  horizontal  line,  is  called  the  axis  of  abscissae,  or 
the  axis  of  x.  The  other  is  called  the  axis  of  ordinates, 
or  the  axis  of  y.  We  shall  speak  of  XOY,  FOX',  X'OY', 
and  Y'OX  as  the  first,  second,  third,  and  fourth  quadrants 
respectively. 

2.  Angles  of  any  Magnitude.  There  are  many  ways  in 
which  a  system  of  coordinates  is  used  in  mathematics.  In 
trigonometry  such  a  system  is  used  primarily  in  defining 

B  1 


2  PLANE    TRIGONOMETRY  [I,  §2 

the  trigonometric  functions,  but  before  we  proceed  to  do  so 
we  shall  extend  our  ideas  of  angles  beyond  the  knowledge 
we  obtained  of  them  in  the  elementary  geometry.  There 
an  angle  is  denned  by  some  such  definition  as  the  following : 
the  plane  figure  formed  by  two  straight  lines  drawn  from 
the  same  point.  The  unit  of  angles  is  either  the  right  angle, 
or  -the  degree,  and  the  largest  angle  usually  dealt  with  is 
equivalent  to  two  right  angles  and  is  often  called  a  straight 
angle.  In  trigonometry,  on  the  other  hand,  we  deal  with 
angles  of  any  magnitude  whatever.  To  do  so  we  introduce 
the  idea  of  motion,  of  revolution.  Starting  from  the  initial 
position  OX,  Fig.  1,  we  may  revolve  the  line  about  0  in 
the  direction  indicated  by  the  arrows,  stopping  in  any 
desired  terminal  position  OPi,  OP2,  OP3,  OP4,  etc.  In  this 
way  angles  of  any  number  of  degrees  whatever  may  be  gen- 
erated. Thus,  if  we  stop  in  the  position  OF,  we  have  an 
angle  of  90°;  in  the  position  OX',  180°;  in  the  position 
OP3,  225° ;  in  the  position  OY',  270°,  and  so  on.  By  mak- 
ing one  whole  revolution  we  should  arrive  at  an  angle  of 
360° ;  two  and  one  half  revolutions,  900°;  etc. 

Not  only  so,  but  we  might  revolve  from  the  initial  posi- 
tion OX  in  the  opposite  direction.  Now  oppositeness  is 
indicated  algebraically  by  the  use  of  the  signs  plus  (+)  and 
minus  (— ).  So  that  if  we  agree  to  take  the  positive  direc- 
tion of  revolution  counterclockwise,  then  clockwise  will 
be  the  negative  direction  and  we  can  thus  generate  negative 
angles  of  any  magnitude  whatever.  Thus,  Fig.  1,  the 
angle  XOP3  is  225°  if  we  have  revolved  in  the  positive 
direction,  but  is  — 135°  if  we  have  revolved  in  the  negative 
direction.  When  an  angle  lies  in  value  between  0°  and  90° 
it  is  said  to  be  an  angle  in  the  first  quadrant  since  its 
terminal  side  lies  in  the  first  quadrant.  An  angle  lying  in 
value  between  90°  and  180°  is  said  to  be  in  the  second 
quadrant;  between  180°  and  270°,  in  the  third  quadrant; 
between  270°  and  360°,  in  the  fourth  quadrant. 


I,  §3] 


TRIGONOMETRIC   FUNCTIONS 


EXAMPLES 
Construct  the  angles 

1.  300°.  3.    750°.  5.    -  1215°. 

2.  -210°.  4.    -495°.  6.    420°. 

Add  the  following  angles  graphically  : 

7.  720°  and  30°.  10.   990°  and  -  60°. 

8.  -  180°  and  60°.  11.    -  45°  and  120°. 

9.  -  90°  and  -  45".  12.    135°  and  -  450°. 

If  A  is  a  positive  angle  in  the  first,  second,  third,  or  fourth  quad- 
rant respectively,  add  graphically 

13.  450°  and  A.  15.    180°  and  -A. 

14.  -  270°  and  A.  16.    -  540°  and  -  A. 

3.   Abscissa,  Ordinate,  and   Distance.     Consider  an  angle, 
positive  or  negative,  of  any  magnitude  whatever*,  XOP,  of 


Fig.  2.  From  P,  any  point  in  the  terminal  side  of  this 
angle,  drop  a  perpendicular  upon  the  axis  of  x.  The  lines 
of  the  figure  are  named  as  follows :  OM  is  called  the 
abscissa  of  the  point  P7  MP  the  ordinate,  and  OP  the  dis- 
tance. The  abscissa  OM  and  the  ordinate  MP  are  together 
called  the  coordinates  of  the  point  P.  Note  very  carefully 
that  the  abscissa  is  always  read  from  0  to  M,  the  ordinate 
from  M  to  P\  that  is,  in  each  case  from  the  axis  to  the 

*  As  a  matter  of  convenience  we  do  not  consider  angles  numerically 
greater  than  360°.  It  is  obvious  that  the  discussion  applies  equally  well 
to  such  angles. 


4  PLANE    TRIGONOMETRY  [I,  §3 

point.  The  distance  is  read  from  0  to  P.  Thus,  for  an 
angle  in  the  second  or  third  quadrant  the  direction  of  the 
abscissa  is  opposite  to  that  of  an  angle  in  the  first  or  fourth 
quadrant.  For  an  angle  in  the  third  or  fourth  quadrant 
the  direction  of  the  ordinate  is  opposite  to  that  of  the 
ordinate  of  an  angle  in  the  first  or  second  quadrant. 
Oppositeness  in  direction  being  distinguished  as  usual  by 
difference  in  algebraic  sign  we  have  the  following  con- 
ventions : 

Tlie  abscissa  measured  to  the  right  of  the  axis  of  y  is 
positive  ;  to  the  left,  negative.  The  ordinate  measured  up- 
ward from  the  axis  of  x  is  positive  ;  downward,  negative. 
The  distance  is  measured  from  the  origin  outward  and  is 
taken  positive. 

EXAMPLES 

1.  The  abscissa  of  a  point  is  3,  its  ordinate  4  ;  find  the  distance. 

2.  The  distance  of  a  point  is  5,  its  ordinate  4  ;  find  the  abscissa. 

3.  The  ordinate  of  a  point  is  —  2,  its  distance  3  ;  find  the  abscissa. 

4.  The  ordinate  of  a  point  is   —  5,   its  abscissa   —  4 ;    find  the 
distance. 

5.  Prove  that  the  square  of  the  distance  of  any  point  is  equal  to 
the  sum  of  the  squares  of  the  abscissa  and  ordinate. 

6.  Prove  that  for  all  points  on  a  straight  line  through  the  origin 
the  ratio  of  the  ordinate  to  the  abscissa  is  constant. 

4.  The  Trigonometric  Functions  Defined.  Let  us  now 
proceed  to  define  the  six  trigonometric  functions  of  an 
angle;  six  quantities  which  depend  upon  the  angle  for 
their  values.  They  are  the  possible  ratios  between  the 
various  pairs  of  the  three  lines  named  in  Art.  3.  Thus, 
Fig.  2,  the 

sine  XOP  =  ordinate  of  P  =  MP       ^ 

^r^^  distance  of  P      OP 

cosine  XOP        =  abscissa  of  P  =™, 
distance  of  P     OP 

tangent  XOP      =  ornate  of  P=MP 
abscissa  of  P     OM 


I,  §5] 


TRIGONOMETRIC   FUNCTIONS 


.' 


cotangent  XOP  =  -;;-•--  ~  -^  ^  = 


abscissa  of  P 

OM 

ordinate  of  P 
distance  of  P 

MP' 
OP 

abscissa  of  P 
distance  of  P 

OM' 

OP 

ordinate  of  P 

MP 

secant  XOP        = 


cosecant  XOP     = 


Three  other  functions  are  sometimes  used :  The  versed 
sine,  which  is  unity  minus  the  cosine ;  the  coversed  sine, 
which  is  unity  minus  the  sine ;  the  suversed  sine,  which  is 
unity  plus  the  cosine.  They  are  relatively  unimportant. 

5.  Trigonometric  Functions  are  Ratios.  The  first  thing 
to  be  noted  about  these  functions  is  that,  being  ratios,  they 
are  independent  of  the  actual  lengths  of  the  abscissa, 


M3 


M2     M, 


I  MO 


FIG.  3. 


ordinate,  and  distance.     Thus,  Fig.  3,  the  triangles 
OM2P.2,  and  OM3P3  being  similar,  their  homologous  sides 
are  proportional,  so  that 


OP,       OP2       OP3  ' 


OM, 


Similarly  the  truth  of  the  statement  may  be  shown  for 
the  remaining  functions. 


6  PLANE    TRIGONOMETRY  [I,  §  6 

6.  Signs  of  the  Functions.  The  second  point  to  be  noted 
is  that  the  signs  of  the  functions  vary  according  .to  the 
quadrant  in  which  the  angle  lies.  Thus,  Fig.  2,  for  the 
angle  XOP  in  the  first  quadrant  the  abscissa,  ordinate  and 
distance  are  all  positive  so  that  all  the  functions  are 
positive.  For  the  angle  XOP  in  the  second  quadrant  the 
ordinate  and  distance  are  positive,  the  abscissa  negative. 
Thus  we  have  for  the  angle  in  the  second  quadrant 


***>?-<£-=. 


The  following  table  gives  the  signs  of  the  functions  in 
the  four  quadrants. 

QUAD.  I  II  III  IV  QUAD. 

sine  -f-          -f  cosecant 

cosine  -f  -f-         secant 

tangent         +  +  cotangent 

EXAMPLES 

Determine  the  algebraic  signs  of 

1.  cos  218°.  3.    sin  1100°.  5.    sec  315°. 

2.  tan  (-460°).         4.    cot  (-  99°).  6.   esc  (- 210°). 

7.   Let  the  student  determine,  as  above,  the  signs  of  the  trigono- 
metric functions  of  angles  in  the  third  and  fourth  quadrants. 


I,  §8] 


TRIGONOMETRIC   FUNCTIONS 


7.  Functions  of  Acute  Angles.  A  special  set  of  definitions 
for  the  functions  of  acute  angles,  which  are  sometimes 
useful  and  should  be  known,  follows  directly  as  a  special 


FIG.  4. 

case  of  the  general  definitions  given  above.     Thus,  Fig.  4, 
in  which  the  angle  XOP  lies  in  a  right  triangle, 

_  ordinate  _  opposite  side 
distance       hypotenuse 

v/-ki>      abscissa      adjacent  side 
cos  XOP  =  —         -  =  -^—  —  > 

distance       hypotenuse 


, 

abs.      adj.  side 


-, 
ord.      opp.  side 


, 

abs.      adj.  side 


ord.      opp.  side 

These  definitions,  it  must  be  noted,  completely  agree 
with  the  more  general  definitions,  but  are  applicable  only 
to  angles  less  than  ninety  degrees,  since  angles  greater 
than  ninety  degrees  cannot  occur  in  right  triangles. 

8.  Reciprocal  Functions.  Two  questions  would  naturally 
suggest  themselves  at  this  point:  Are  the  trigonometric 
functions  of  an  angle  related  to  each  other  in  any  particular 


8  PLANE    TRIGONOMETRY  [I,  §8 

way?  and,  second,  if  there  be  a  definite  relation  between 
two.  given  angles  will  the  functions  of  those  angles  bear 
some  special  relation  to  each  other?  We  shall  proceed  to 
answer  the  first  of  these  questions  affirmatively,  but  shall 
leave  the  discussion  of  the  second  question  to  a  later 
chapter  (Chap.  II).  Thus,  if  a  be  any  angle,  it  follows 
by  the  definitions  of  the  trigonometric  functions  that 

ordinate 
sin  «  =  YT— 

listance 


~~ 


tana  = 


ord. 
abscissa         1 


distance  ~~  dist.  ~~  sec  n 
abs. 

ordinate  _     1  1 

abscissa  ~~  abs.  ~~  cot  a9 
ord. 


or,  the  sine  and  cosecant,  the  cosine  and  secant,  the  tangent 
and  cotangent  respectively  of  the  same  angle  are  reciprocals 
of  each  other. 

9.   Tangent,  Sine  and  Cosine.     Again,  by  definition,  and  by 

Art.  8, 

ordinate 
ordinate      distance      sin  a 


cot  a  = 


abscissa      abscissa     cos  a' 
distance 

1         cos  a 


tan  a     sin  a     sin  a 
cos  a 


These  relations  may  be  proved  otherwise,  thus,  Fig.  5 : 

MP 
_„.„      MP      OP      sin  XOP 

tan  XOP  =  T^TT  =  T^TT  =  v/^p> 

OM      OM     cos  XOP 
~OP 


I,  §11] 


TRIGONOMETRIC   FUNCTIONS 

QM 

OP  _  cos  XOP 
MP~  sin  XOP' 
OP 


9 


_ 
~         " 


y 
FIG.  5. 


10.    Sine  and  Cosine.     Also,  Fig.  5,  it  is  obvious  that 

MP2  +  OM2  =  OP2. 
Dividing  each  term  by  OP2  gives 


OP) 
Whence,  by  definition, 

(sin  XOP)2+(cos  XOP)2  =  1 
or,  as  it  is  usually  written,  letting  a  =  Z  XOP, 
sin2  a  +  cos2  a  =  1. 

11.   Tangent   and    Secant.     Similarly,   writing    the    first 
equation  of  Art.  10  in  the  form 


and  dividing  each  term  by  OM 2,  we  have 
(OPV_(MP\* 

(OM)-(OM) 


10  PLANE   TRIGONOMETRY  [I,  §11 

That  is  (sec  XOP)2  =  (tan  XOPy  +  1 

or,  sec2  a  =  tan2  a  +  1. 

In  the  same  way  we  obtain  the  relation 
csc2  a  =  cot2  a  +  1. 

12.  Fundamental  Relations.     These  relations,  summarized 
below,  are  of  great  importance  and  must  be  memorized. 

sin  a  = 
csc  a  = 

tan  a  =  ^^,    cot  a  =  ZZL=. 
cos  a  sin  a 

sin2  a  +  cos2  a  =  1.  (3) 

sec2  a  =  1  4-  tan2  a,        csc2  a  =  1  4-  cot2  a.  (4) 

13.  By  means  of  the  identities  of  Art.  12  the  value  of 
any  one  of  the  trigonometric  functions  may  be  expressed  in 
terms  of  each  of  the  other  five.     Thus,  by  (3) 


1 

cos  a-    l 

tflfi  n  — 

csc  a 

sec  a' 

cot  a 

sin  a' 
sina 

cosa 
<-.ota.-c°8a 

tan  a 

sin  a  =  VI  —  cos2  a. 
By  (2),  (1),  and  (4), 

Bina  =  tana- cos  «=^=s-    tan  a 


sec  a      VI  4-  tan2  a 
where  the  radical  may  be  either  plus  or  minus. 

14.  To  Compute  the  Values  of  the  Other  Functions  when  One 
Function  of  an  Angle  is  Given.  By  means  of  the  relations 
of  the  preceding  article  if  the  value  of  any  one  function  of 
an  angle  be  given,  the  values  of  the  remaining  functions 
may  be  found,  but  a  simpler  method  of  obtaining  them  is 
illustrated  by  the  following  examples. 


I,  §  14] 


TRIGONOMETRIC   FUNCTIONS 


11 


Example  1.     Given  sin  A  =  —  -f  ,  find  the  values  of  the 
remaining  functions. 

ord.          m-2         2^   -2 
33* 


sin  A  = 


dist. 


m  •  3 


The  distance  being  always  positive,  the  minus  sign  nec- 
essarily is  taken  with  the  ordinate.  Therefore,  Fig.  6, 
construct  an  angle  whose  ordinate  is  —  2  and  whose  dis- 


FIG.  6. 


tance  is  3,  or  any  multiple  (m)  of  —  2  and  3.  The  third 
side  of  the  right  triangle  is  ±  V9  —  4  =  ±  V5,  This  is 
the  value  of  the  abscissa  and  we  may  write  the  values  of 
the  six  functions  from  the  definitions. 


, 

o 


sn 


r, 

V5 


2 

-7= 

v5 


V5 


sec  XOP2  =  — , 


12 


PLANE   TRIGONOMETRY 


[I,  §  14 


Example  2.     Given  cot  A  =  —  -f ,  find  the  remaining  func- 


tions. 


cot  A  = 


abs. 
ord. 


m  •  5 
m  •  4 


5 
-4' 


since  either  the  abscissa  or  the  ordinate  may  be  negative. 

Construct  an  angle  XOP1}  Fig.  7,  having  an  abscissa  —  5 

and  an  ordinate  -f  4,  and  an  angle  XOP2  having  an  abscissa 


M,-5  xx 


+  5 


Wf 


y' 
FIG.  7. 


4-  5  and  an  ordinate  —  4.     In  each  case   the  distance  is 
found  to  be  V25  +  16  =  V41,  and  we  may  write 


sn 


4 

V4l' 


sin  XOP2  =  - 


V4l' 


sec 


V41 

5' 
5 

4' 
V41 


cos  XOPo  = 


5 


2  —  —  := 


cot 


sec 


o  = 


V41 


_5 
4' 

VH 


esc  XOPl  = 


V41 


esc  XOP2  =  — 


V41 


It  will  be  seen  that  the  ambiguity  of  the  two  sets  of  values 
will  occur  in  every  case,  no  matter  what  function  be  given  and 
no  matter  whether  the  sign  of  the  given  function  be  plus  or 
minus. 


I,  §  14]  TRIGONOMETRIC   FUNCTIONS  13 

EXAMPLES 

Find  the  values  of  the  remaining  functions,  given  that 

1.  sin  a  =  J.  3.   cot  a  =  -  3.  5.   sec  a  =  4. 

2.  cos  a  =  -  f .  4.   tan  a  =  f .  6.   esc  a  =  —  ^. 

7.  If  sin  x  =  5,  can  the  values  of  the  remaining  functions  be 
found  ?    Why  ? 

8.  If  sec  x  =  |,  can  the  values  of  the  remaining  functions  be 
found  ?     Why  ? 

9.  If  tan  x  =  —  4,  can  the  values  of  the  remaining  functions  be 
found  ?     Why  ? 

"  10.    Given  sec  a  =  |,  find  the  functions  of  90°  —  a. 
11.    Given  cot  a  =  x,  find  the  functions  of  90°  —  a. 


Prove  the  following  relations  : 
12.   cos  a  =  - 


19 


13.    cota  =  —  20        coscc 


tana 


Vsec2  a  —  1  i  —  sin  a        cos  a 


14.   seCtt  =  -  ggg-g  --  21. 


. 

Vcsc2  a  —  1  sec  a 


=  sin  a. 


15.   sin  «=  -  22. 

Vl  +  cot2  a  sin2  a 


16.  tanft  =        -.  23.  =  cos  «. 

cos  a  esc  a 

17.  tan  a  =       sin  ff    - .  24.   tan  a  •  esc  a  =  sec  a. 

Vl  -  sin2  a 

18.  cot  a  =  — cos  a  25.   cos  a-  sec  a  = . 

vl  —  cos2  a  cos  ot  •  sec  a 

26.   sec2  a  —  csc2a  =  tan2  a  -  cot2  a. 


CHAPTER   II 


IDENTICAL  RELATIONS  AMONG  THE  FUNCTIONS  OF 
RELATED  ANGLES.  THE  VALUES  OF  THE  FUNC- 
TIONS OF  CERTAIN  ANGLES 

15.  Functions  of  Negative  Angles.  We  shall  now  proceed 
to  determine  the  relations  which  exist  among  the  functions 
of  two  angles  when  those  angles  are  related  in  some  par- 


FIG.  8. 

ticular  way.  Let  us  consider  first  two  angles  one  of  which 
is  the  negative  of  the  other,  Fig.  8.  Let  the  value  of  the 
positive  angle  XOP  be  a,  and  of  the  numerically  equal 
negative  angle  XOP '  be  —  a.  On  the  terminal  sides  of 
these  angles  lay  off  the  equal  distances  OP  and  OP',  and 
drop  perpendiculars  from  P  and  P1  upon  the  axis  of  x. 
These  perpendiculars  will  obviously  cut  the  axis  of  a;  in 

14 


II,  §  16]  IDENTICAL  RELATIONS  15 

the  same  point,  M,  and  the  two  right  triangles  MOP  and 
MOP'  will  be  congruent.  Therefore,  OF'=OP,  OM=  OM, 
and  MP'  =  -  MP.  We  then  have 


MP'      -  MP 


Thus  any  function  of  a  negative  angle  is  equal,  numeri- 
cally, to  the  same  function  of  an  equal  positive  angle.  The 
algebraic  sign  is  determined  by  the  quadrant  which  —  a  lies 
in  when  a  is  acute. 

16.  Functions  of  90°  —  a.  Consider  next  two  angles, 
a  and  90°  -  «,  Fig.  9.  Let  XOP  be  the  angle  a  and  XOP' 
be  90°  —  a.  Lay  off  on  the  terminal  sides  of  these  angles 
the  equal  distances  OP  and  OP  ',  and  from  P  and  P  '  drop 
perpendiculars  PM  and  P'M1  upon  the  axis  of  x.  Then 
obviously  the  right  triangles  MOP  and  M'OP'  are  congru- 
ent, and  OF  =  OP,  M'P  '  =  OM,  and  OM  '  =  MP.  There- 
fore, 


1 


*  By  virtue  of  Art.  12,  it  is  necessary  to  memorize  only  sin  (—  a)  and 
cos  (—a). 


16 


PLANE   TRIGONOMETRY 


[II,  §  16 


Sec(90°-a)  =  |j;  =  £|=csc«, 


°BC  (<W- «)=-£!;=  ££=  see 


a. 


y 

\ 

yjP                                                                 P 

\ 

y 

x' 

/90°-  v^\ 

\ 

^« 

O 

M'         M                                         MO 

y' 

y' 

y 

y 

/ 

xf     M             'M' 

'"N                                p' 

o  %,           xx-  T  —  :  —  : 

a 

rt~~ 

5>  M                      x 

/ 

^-a                                                9Q,_ 

1 

p' 

y' 

y'Np 

FIG.  9. 

Thus  we  see  that  each  of  the  functions  of  90°  —  a  is 
equal,  numerically,  to  the  co-function  of  the  angle  a.  The 
important  case  is  when  a  is  acute,  and  a  and  90°  —  a  are 
complementary  angles.  Indeed  it  was  because  of  this  rela- 
tion that  the  cosine,  cotangent,  and  cosecant  received  their 
names.  They  are  the  sine,  tangent,  and  secant  of  the 
complementary  angle. 

17.  Functions  of  90°  +  a.  We  shall  consider  two  more 
cases,  limiting  the  discussion  to  acute  values  of  a,  although 


II,  §  18] 


IDENTICAL   RELATIONS 


17 


the  results  will  be  equally  true  for  any  value  of  a  whatever. 
In  Fig.  10  let  the  angle  XOP  be  a  and  XOP'  be  90°  +  «. 
On  the  terminal  sides  of  these  angles  lay  off  the  equal  dis- 
tances OP  and  OP',  and  from  P  and  P'  drop  perpendiculars 
PM  and  P'M '  upon  the  axis  of  x.  It  follows  that  the  two 


FIG.  10. 


triangles  MOP  and  M'OPf  are  congruent  and  that   OP7 
=  OP,  M'P'  =  OM,  OM'  =  -  MP.     Therefore, 


18.   Functions  of  180°  4-  a. 
be  a  and  XOP'  be  180°  +  a. 


In  Fig.  11  let  the  angle  XOP 
On  the  terminal  sides  of  these 


18 


PLANE   TRIGONOMETRY 


[11,118 


angles  lay  off  the  equal  distances  OP  and  OP,  and  from 
P  and  P  drop  perpendiculars  PM  and  P'M'  upon  the  axis 


x'    M 


of  X.    Then  the  triangles  MOP  and  M'  OP'  are  congruent 
and  OP'  =  OP,  M'P'  =  -  MP,  OM'=  -  OM.     Therefore, 


sin  (180°  +  «)  = 


=~ 


19.  Generalization.  In  a  similar  manner  may  be  found 
analogous  relations  connecting  the  functions  of  an  angle  a 
with  the  functions  of  any  integral  multiple  of  90°  plus  or 
minus  a.  Upon  examining  these  relations  we  are  led,  by 
iduction,  to  express  them  in  the  following  general  rule. 

Any  function  of  an  even  *  multiple  of  90°  plus  or  minus  a 
is  the  same  function  of  the  angle  a. 

*  Zero  is  taken  as  an  even  number,  so  that  the  rule  includes  the  case 
of  Art.  15. 


II,  §  19]  IDENTICAL   RELATIONS  19 

Any  function  of  an  odd  multiple  of  90°  plus  or  minus  a  is 
the  co-function  of  the  angle  a. 

The  algebraic  sign  of  the  value  is  determined  by  the  quad- 
rant (counting  in  the  positive  direction)  in  which  the  terminal 
side  of  the  angle  lies  when  a  is  acute. 

Examples. 

1.  sin  (720°  —  «)  =  —  sin  a,  since  720°  is  an  even  multiple 
of  90°  and  the  terminal  side  of  720°  —  a,  when  a  is  acute, 
lies  in  the  fourth  quadrant. 

2.  cot  (—  90°  —  «)=  tan  a,  since  —  90°  is  an  odd  multiple 
of  90°  and  the  terminal  side  of  —  90°  —  a,  when  a  is  acute, 
lies  in  the  third  quadrant. 

3.  sec  (—  180°  +  a)—  —  sec  a,   since    —180°   is   an   even 
multiple  of  90°  and  the  terminal  side  of  —  180°  -f  a,  a  acute, 
lies  in  the  third  quadrant. 

4.  tan  281°  =  tan  (270°  +  11°)  =  -  cot  11°,  or 
tan  281°  =  tan  (360°  -  79°)=  -  tan  79°, 

since  I  ,™0  is  an  j  multiple  of  90°  and  the  terminal 

side  of  281°  lies  in  the  fourth  quadrant. 

(~z  & 

EXAMPLES 

By  means  of  a  geometrical  construction  express  each  of  the  follow- 
ing as  a  function  of  a,  where  a  is  an  acute  angle.  Check  your  results 
by  the  rule  given  above. 

1.  cos  (270°  +  a}.  5.  cot  (270°  -  a). 

2.  sin  (180°  -a).  6.  sec  (270°  -  a). 

3.  esc  (-  90°  +  a).  7.  sin  (-  180°  -  a). 

4.  tan  (540°  + a).  8.  cos(- 270°  +  a). 

Express  as  a  function  of  an  acute  angle 

&  9.    sin  324°.  13.  sec  (-537°). 

10.  cos  (-375°).  14.  cot  1140°. 

11.  tan  457°.  15.  tan  496°. 

12.  esc  (-801°  32').  16.  cos  (-480°). 


20 


PLANE    TRIGONOMETRY 


[II,  §20 


20.  Functions  of  Certain  Angles.     We  see  by  the  preced- 
ing article  that  the  functions  of  angles  greater  than  90°, 
and  of  negative  angles,  can  be  expressed  in  terms  of  the 
functions  of  angles  lying  between  0°  and  90°.     It  follows 
that  if  we  wish  to  use  the  trigonometric  functions  for  com- 
putation or  for  other  purposes  we  need  find  their  values 
only  for  all  positive  acute  angles.     We  shall  not  discuss 
the  methods  by  means  of  which  these  values  are  computed 
in  general,  but  shall  proceed  to  find  the  values  of  the  func- 
tions of  certain  angles  which  frequently  occur.     We  shall 
then,  in  the  following  chapter,  show  how  we  may  find  the 
values  of  the  functions  of  any  angle  from  tables  with  which 
we  are  provided.     We  shall  see,  also,  how  the  values  thus 
found  may  be  used  in  the  solution  of  triangles ;  that  is,  in 
finding  the  unknown  parts,  angles  or  sides,  of  a  triangle 
from  parts  which  are  given. 

21.  Functions  of  30°  and  60°.     Let  the  angle  XOP,  Fig.  12, 
be  an  angle  of  30°,  and  from  P  drop  a  perpendicular,  PM, 


aVT 


FIG.  12. 


upon  the  axis  of  x.     Then,  as  we  know,  the  angle  0PM  is 


a 


60°,  and  if  OP  have  the  value  a,  MP  must  be  equal  to  - 


and  OM  equal  to 


aV3 


Therefore,  by  definition, 


II,  §221 


IDENTICAL   RELATIONS 


21 


tan  30°  = 


a 

MP     ~~2 


OM     aV3      V3 


oV3 

03f        2         V3 
cos30  =         =  — =  ^T 

oV3 


V3 


MP     a 

2 


By  a  similar  construction,  or  by  the  relations  of  Art.  16, 
the  following  values  may  be  derived  : 


sec  60°  =  2, 


sin  60°  =  ^2,  tan60°=V3, 

2 


cot  60°  = 


V3 


esc  60°  = 


vs 


FIG.  13. 

22.  Functions  of  45°.  Let  the  angle  XOP,  Fig.  13,  be  an 
angle  of  45°,  and  from  P  drop  a  perpendicular,  PM,  upon 
the  axis  of  x.  Then  the  angle  0PM  is  an  angle  of  45°,  and 
if  OM  have  the  value  a,  MP  also  will  be  equal  to  a  and  OP 
will  be  aV2.  Therefore,  by  definition, 


22  PLANE   TRIGONOMETRY  [II,  §22 


5 

aV2      V2  OP      aV2 


23.  Functions  of  Other  Angles  Readily  Found.  By  sim- 
ilar constructions  the  functions  of  120°,  150°,  135°,  etc.,  or, 
in  general,  any  integral  multiple  of  90°  plus  or  minus  30°, 
60°,  or  45°,  may  be  found.  They  may  be  found  more 
readily,  however,  by  using  the  rule  given  in  Art.  19.  Thus, 

sin  120°  =  sin  (90°  +  30°)  =  cos  30°  =  — 

.  2 


or 


sin  120°  =  sin  (180°  -  60°)  =  sin  60°  =        . 


EXAMPLES 
Find  the  values  of  the  functions  of 

1.  120°.  4.    210°.  7.   300°. 

2.  135°.  5.    225°.  8.   315°. 

3.  150°.  .  6.   240°.  9.    330°. 

Prove  that 

10.  sin  210°  tan  300°  =  sin  120°. 

11.  sec  315°  sec  300°  =  sec  240°  sec  225°. 

12.  tan  210°  :  cos  150°  =  tan  150°  :  cos  330°. 

13.  esc  330°  sec  315°  sin  225°  =  -  sec  120°. 

24.  Functions  of  Zero.  Let  the  value  of  the  angle  XOP, 
Fig.  14,  be  represented  by  a,  and  from  P,  any  point  in  the 
terminal  side  of  the  angle,  drop  a  perpendicular,  PM,  upon 
the  axis  of  x.  By  definition, 


II,  §24] 


IDENTICAL   RELATIONS 


23 


Now,  for  the  sake  of  convenience  keeping  the  distance 
OP  constant  in  length,  let  the  line  OP  approach  nearer  and 
nearer  to  the  position  OX.  Then  the  angle  a  can  be  made,* 
smaller  than  any  angle  that  may  be  assigned,  however 


FIG.  14. 

small,  or,  as  it  is  otherwise  expressed,  a  will  approach  the 
limit  zero.     At  the  same  time  MP  will  approach  zero  as  a 

limit,  and  OM  will  approach  OP  as  a  limit. 

OM 


Then  will 


approach  the  limit  zero  and 


OP 


OP 

will  approach  the  limit 


unity.  Thus,  as  the  angle  approaches  the  limit  zero  (or, 
becomes  smaller  than  any  value  that  may  be  assigned,  how- 
ever small)  its  sine  approaches  the  limit  zero  (or,  becomes 
smaller  than  any  value  that  may  be  assigned,  however 
small)  and  its  cosine  approaches  the  limit  unity  (or,  differs 
from  unity  by  a  number  smaller  than  any  number  that  may 
be  assigned,  however  small).  This  may  be  written 

limit  sin  a  =  0, 


limit  cos  a  =  1. 

o=0 


OP 


Again,  by  definition,  esc  a  = ,  and  as  a  grows  smaller 

MP  Qp 

OP  remains  constant  and  MP  grows  smaller,  so  that  

becomes  continually  greater.     Finally,  when  a  approaches 
zero  as  a  limit,  MP  becomes  smaller  than  any  number  that 

*  And  will  remain. 


1*5 


24 


PLANE    TRIGONOMETRY 


[II,  §24 


OP 

may  be  assigned,  however  small,  and  -  becomes  greater 

than  any  number  that  may  be   assigned,  however   great. 

OP 

This  we  express  by  saying  that  -  approaches  the  limit 

infinity,  or  increases  without  limit.     We  may  then  write 
limit  esc  a  =  00. 

a=0 

Similarly  it  may  be  shown  that 
limit  tan  a  =  0,  limit  cot  a  =  oo,  limit  sec  a  =  1. 


These  relations  are  often  briefly  expressed, 


sin  0°  =  0, 
cosO°  =  l 


tan  0°  =  0, 
cot  0°=  oo, 


sec  0°  =  1, 
esc  0°=  co. 


(6) 


to  which  there  is  no  objection  if  we  remember  that  these  are 
merely  abbreviations  of  the  preceding  statements,  and  that 


FIG.  15. 


0°  means,  not  that  we  have  no  angle,  but  that"  we  are  deal- 
ing with  an  angle  which  becomes  smaller  than  any  value 
that  may  be  assigned,  however  small ;  and  that  when  this 
happens  the  sine  of  the  angle  also  becomes  smaller  than 
any  value  that  may  be  assigned,  however  small,  the  cotangent 


II,  §26]  IDENTICAL   RELATIONS  25 

becomes  greater  than  any  value  that  may  be  assigned,  how- 
ever great,  the  cosine  approaches  the  limit  unity,  etc. 

25.  Functions  of  90°.  Let  the  angle  XOP,  Fig.  15,  be 
represented  by  a,  and  let  OJf,  MP,  and  OP  be  respectively 
the  abscissa,  ordinate,  and  distance  of  P.  Also,  keeping  the 
distance  OP  constant,  let  the  line  OP  approach  0  Y  as  its 
limiting  position.  Then, 

a  approaches  the  limit  90°, 
OM  approaches  the  limit  zero, 
MP  approaches  the  limit  OP. 

Therefore, 

limit  sin  a  =  limit  -  =  1. 

a=90°  OP 

limit  cos  a  =  limit  -  =  0. 

a=90°  OP 

limit  tan  a  =  limit  -  =  oo, 

a=90°  OM 

limit  cot  a  =  limit  -  =  0. 

MP 


OP 

limit  sec  a  =  limit  -  —  =  oo, 

a=90°  OM 

OP 

limit  esc  a  =  limit  -  =  1. 

a=90°  MP 

With  the  same  understanding  as  in  the  preceding  article 
these  may  be  written 

sin  90°  =  1,  tan  90°  =  oo,  sec  90°  =  oo, 

cos  90°  =  0,  '          cot  90°  =  0,  esc  90°  =  1. 

26.    The  student  should  find,  as  in  Arts.  24  and  25,  the 
following  : 

sin  180°  =  0,  tan  180°  =  0,        sec  180°  =  -  1, 

cos  180°  =  -  1,        cot  180°  =  oo,       esc  180°  =  oo. 

sin  270°  =  -  1,        tan  270°  =  oo,       sec  270°  =  oo, 
cos  270°  =  0,  cot  270°  =  0,       esc  270°  =  -  1. 


26 


PLANE    TRIGONOMETRY 


[II,  §27 


27.  Limiting  Values  of  the  Functions.  We  have  seen 
(Art.  20)  that  all  possible  numerical  values  of  the  trig- 
onometric functions  are  given  by  angles  lying  between  0° 
and  90°.  Let  us  now  see  between  what  limits  the  values  of 
the  functions  lie.  From  the  discussion  and  figures  of  articles 
24  to  26  we  see  that 

the  sine  and  cosine  of  an  angle  lie  between  —  1  and  +  1, 
the  tangent  and  cotangent  lie  between  —  oo  and  -f-  oo, 
the  secant  and  cosecant  lie  between  1  and  oo  or  between  —  1 
and  —oo. 

It  is  well  to  note  also,  for  angles  in  the  first  quadrant, 
that  as  the  angle  increases  the  direct  functions  increase, 
the  co-functions  decrease. 

A  very  convenient  and  simple  way  to  remember  the  range 
of  values  and  the  signs  of  the  trigonometric  functions  is  by 


FIG.  16. 

means  of  the  unit  circle,  a  circle  with  unit  radius,  which 
need  not  be  actually  drawn  but  merely  visualized.  Draw 
such  a  circle,  Fig.  16,  with  its  center  at  the  origin  of  co- 
ordinates, and  let  XOP  be  any  angle.  Drop  the  perpendicu- 
lar PM  upon  the  axis  of  X,  and  draw  LQ  tangent  to  the 
circle  at  L  and  meeting  OP  produced  in  Q.  Then,  by 
definition, 


II,  §27]  IDENTICAL   RELATIONS  27 


tan  XOP  =        =        =  ±  LQ,  etc. 
OL      ±1 

If  now  the  line  OP  be  pictured  as  revolving  from  the 
position  OL,  the  sine  of  the  angle  XOP,  namely  MP,  will 
be  seen  to  increase  from  zero  and  approach  unity  as  the 
angle  approaches  90°.  The  cosine,  namely  OM,  decreases 
from  unity  to  zero,  and  the  tangent  (LQ)  increases  without 
limit.  Also,  as  the  angle  increases  beyond  90°,  the  direc- 
tions of  the  lines  MP  and  OM  indicate  the  signs  of  the 
sine  and  cosine.  The  other  functions  follow  directly  from 
these  two  by  virtue  of  the  relations  of  Art.  12. 


CHAPTER   III 


THE   SOLUTION   OF    RIGHT   TRIANGLES.    LOGARITHMS 
AND  COMPUTATION  BY  MEANS  OF  LOGARITHMS 

28.  Solution  of  Right  Triangles.  With  the  definitions  of 
the  trigonometric  functions  and  tables  giving  their  nu- 
merical values  we  are  now  prepared  to  solve  right  tri- 
angles ;  that  is,  to  find  the  values  of  the  unknown  parts 
from  those  that  are  known.  Two  parts  in  addition  to  the 

right  angle  must  be  known, 
and  one  at  least  of  these  parts 
must  be  a  side.  We  have  then 
the  general  rule  of  procedure  : 
Select  that  trigonometric  func- 
tion which  involves  the  two 
known  parts  and  one  unknown 
FlG-  17'  part.  The  value  of  the  un- 

known part  can  then  be  computed  by  elementary  algebraic 
processes. 

Example  1.  Given  A  =  32°  16',  a  =  124,  C  =  90°,  find  B, 
&,  and  c.  See  Fig.  17. 

Obviously  B  =  90°  -  A  =  90°  -  32°  16'  =  57°  44'.     Then 


a  =124 


A-32'16' 


:  =  90° 


cot  A  =  -, 
a 


A      a 

sin  A  =  -, 
c 


or 


b  =  a  cot  A. 

From  the  tables  we  find 
cot  A  =  1.5839. 


a 


sin  A 


sin  A  =  .5338. 


28 


Ill,  §28]    SOLUTION   OF   RIGHT   TRIANGLES  29 

Therefore, 

6  =  124x1.5839  c=  124 


.5338 
=  196.4.  =  232.3. 

Example  2.     Given  a  =  50,  6  =  60,  C  =  90°,  find  A,  B, 
and  c. 

In  this  case, 


A  =  39°  48'. 

B  =  90°  -  A  =  50°  12'. 


To  find  c  we  may  use  either 


sin  A  =  -  or  c2  =  a2  -f 


sin  A 
50  =V2500  +  3600 


.6402 
=  78.1.  =  78.1. 

EXAMPLES 
Solve  the  following  right  triangles  : 


1.    a  =  250, 

3.    a  =  .55, 

5.   ^4  =  59°  58', 

A  <=  36°  22'. 

c  =  .70. 

b  =  412. 

2.   a  =  37.5, 

4.   B  =  72°  6', 

6.    £  =  24°  33', 

6  =  40.1. 

c  =  502. 

a  =  211. 

PROBLEMS 

7.  What  is  the  height  of  a  flagpole  if  at  a  horizontal  distance  of 
200  feet  from  the  foot  of  the  pole  the  angle  of  elevation  of  its  top  is 
19°  28'  ? 

8.  A  rope  is  stretched  taut  from  the  top  of   a  building  to   the 
ground,  and  is  found  to  make  an  angle  of  58°  56'  with  the  horizontal. 
If  the  building  is  61  feet  high  how  long  is  the  rope  ? 


30  PLANE   TRIGONOMETRY  [III,  §28 

9.  If  a  tree  74.3  feet  high  casts  a  shadow  42.6  feet  long,  how 
many  degrees  above  the  horizon  is  the  sun  ? 

10.  A  man  walking  on  level  ground  finds,  at  a  certain  point,  that 
the  angle  of  elevation  of  the  top  of  a  tower  is  30°.     He  walks  directly 
toward  the  tower  for  a  distance  of  300  feet  and  then  finds  the  angle 
of  elevation  of  the  top  to  be  60°.     What  is  the  height  of  the  tower  ? 

11.  At  a  point,  J.,  south  of  a  tower  the  angle  of  elevation  of  the 
top  of  the  tower  is  60°.     At  another  point  300  feet  east  of  A  the  angle 
of  elevation  is  30°.     What  is  the  height  of  the  tower  ? 

12.  The  angles  of  a  right  triangle  are  42°  and  48°  ;  the  hypotenuse 
is  200  feet.     What  is  the  length  of  the  perpendicular  from  the  right 
angle  to  the  hypotenuse  ? 

13.  The  height  of  a  gable  roof  is  20  feet,  its  width  42  feet.     What 
is  the  pitch  of    the  roof  ;     that  is,   the  angle  it  makes  with  the 
horizontal  ? 

14.  From  where  I  stand  a  tree  50  feet  away  has  an  angle  of  eleva- 
tion of  43°  31'.     From  the  same  point  another  tree,  75  feet  distant, 
has  an  angle  of  elevation  of  32°  20'.     Which  tree  is  the  taller  and  by 
how  much  ? 

29.  Logarithms.  The  solution  of  right  triangles  as  thus 
explained  is  simple  in  theory  but  may  become  laborious  in 
practice  because  of  the  arithmetic  computation  involved. 
Fortunately  we  have  in  logarithms  a  device  for  simplifying 
such  computation.  The  base  of  a  system  of  logarithms  is, 
in  general,  any  arbitrarily  chosen  number.*  In  practice 
two  systems  are  used :  the  Briggsian  or  common  system  of 
which  the  base  is  10 ;  and  the  Napierian  system  of  which 
the  base  is  e  =  2.718  •••.  The  logarithm  of  a  number  to  a 
given  base  (a)  is  the  exponent  of  the  power  to  which  the 
base  (a)  must  be  raised  to  produce  the  number.  Thus,  if 
ax  =  m,  then  x  is  the  logarithm  of  ra  to  the  base  a ;  written 
x  =  loga  m. 

The  word  power  is  used  here  in  its  broader  sense  to  in- 
clude fractional  and  negative  exponents.  Defining  frac- 
tional and  negative  exponents  in  such  a  way  that  the  laws 
of  exponents  —  aman  =  am+n ;  (am)n  =  amn  —  hold  for  nega- 
tive numbers  and  fractions  as  well  as  for  positive  integers, 

*  Some  numbers,  unity,  for  example,  cannot  be  so  used. 


Ill,  §30]    SOLUTION   OF   RIGHT   TRIANGLES  31 

values  of  x  may  be  found  to  satisfy,  approximately  at  least, 
such  an  equation  as  ax  =  &,  no  matter  what  values  a  and  b 
may  have.  Thus,  given  any  number,  a,  by  raising  it  to  a 
suitable  power,  p,  and  extracting  a  suitable  root,  g,  of  the 
result,  we  can  obtain  any  other  number,  b  ;  that  is,  -^/~ap  =  b. 

But  this  may  be  written  a9  =  b  or  ax  =  6,  where  #  =  -^ ,  the 

division  of  p  by  q  being  carried  out  to  any  desired  number 
of  decimal  places.  We  then  call  x  the  logarithm  of  b  to  the 
base  a. 

30.  The  Common  System.  For  purposes  of  computation 
the  common  system,  base  10,  is  used.  Let  us  form  a  table 
of  powers  of  10  and  express  the  relations  in  terms  of 
logarithms. 

10-s  =  .001,  or  logw  .001  =  -  3. 

10-2  =  .01,  Iog10.01 

10-1  =  .l,  loglo.l 

100  =i9  loglol        =o. 

101  =10,  Iog1010      =1. 

102  =  100,  Iog10 100    =  2. 

103  =1000,  Iog10  1000  =  3,* 
etc.  etc. 

This  table  could  be  extended  indefinitely  in  either  direc- 
tion. If  we  examine  the  table  we  notice  that  to  produce  a 
number  between  1  and  10  we  must  raise  the  base  10  to  a 
positive  power  between  0  and  1 ;  to  produce  a  number  be- 
tween 10  and  100,  the  exponent  of  the  base  must  lie  between 
1  and  2  ;  for  a  number  between  100  and  1000,  the  exponent 
must  lie  between  2  and  3,  and  so  on.  In  other  words,  the 
logarithm  of  a  number  between  1  and  10  lies  between  0  and 
1,  and  is,  therefore,  a  fraction,  always  expressed  as  a  deci- 
mal. The  logarithm  of  a  number  between  10  and  100  lies 
between  1  and  2,  or  is  1  plus  a  decimal.  The  logarithm  of 

*  Hereafter  in  this  work  we  shall  not  write  the  base  10.  Thus  log  7 
means  Iog107.  In  general,  however,  except  in  works  on  trigonometry,  if 
no  base  is  written,  e  —  2.718  -"is  understood. 


32  PLANE    TRIGONOMETRY  [III,  §  30 

a  number  between  100  and  1000  is  2  plus  a  decimal.  The 
logarithm  of  a  number  is  thus  seen  to  consist,  in  general, 
of  two  parts,  an  integral  part  and  a  decimal  part.  The 
integral  part  is  called  the  characteristic  of  the  logarithm  ;  the 
decimal  part  is  called  the  mantissa.  The  results  of  our 
observations  may  be  summarized  thus  : 


NTTMBER  BETWEEN  CONTAINS 

1  and  10  1  integral  digit  0 

10  and  100  2  integral  digits  1 

100  and  1000  3  integral  digits  2 


Whence  we  formulate  the  law :  TJie  characteristic  of  the 
logarithm  of  a  number  is  one  less  than  the  number  of  digits 
in  the  integral  part  of  the  number. 

On  the  other  hand,  we  observe  from  the  table  of  this 
article  that  if  a  number  contain  no  integral  digits,  that  is, 
if  it  be  purely  decimal,  its  logarithm  is  negative.  The 
characteristic  in  this  case  can  be  got  by  counting  the  num- 
ber of  zeros  before  the  first  significant  figure,  prefixing  the 
minus  sign.  It  is  usual,  and  better,  however,  except  for 
special  purposes,  not  to  write  the  characteristic  of  the  loga- 
rithm of  a  decimal  number  in  the  form  just  stated,  for 
reasons  which  will  now  be  pointed  out. 

31.  The  Mantissa.  In  the  common  system  the  mantissa 
of  the  logarithm  of  a  number  can  be  made  to  depend  only 
upon  the  sequence  of  digits  in  the  number,  and  be  inde- 
pendent of  the  position  of  the  decimal  point.  Let  us 
assume  that  we  know  the  logarithm  of  1.285  to  be  0.1089. 
It  follows,  multiplying  successively  by  ten,  that 

10o.io89  =  L285,         or        log  1.285  =  0.1089. 

101-1089  =  12.85,  log  12.85  =1.1089. 

102-1089  =  128.5,  log  128.5  =2.1089. 

103.io89  =  1285,  log  1285    =3.1089. 

104.io89  =  1285o,  log  12850  =  4.1089. 


Ill,  §31]    SOLUTION   OF   RIGHT   TRIANGLES  33 

which  verifies  the  law  we  have  stated.     If,  however,  we 
divide  10°-1089  successively  by  10  we  find 

10o.io89-i  =  10-o.89ii  =  1285,         or        log  .1285     =  -  0.8911. 
10-i.89ii  =  .01285,  log  .01285  =  -  1.8911. 

10-2.89ii  =  .001285,  log  .001285  =  -  2.8911. 


This  is  the  true  form  of  the  logarithm  of  a  purely 
decimal  number,  and  for  certain  purposes  this  is  the  form 
which  must  be  used.* 

It  is  obvious  from  the  preceding  discussion  that  the 
mantissa  corresponding  to  a  given  sequence  of  digits  re- 
mains the  same  as  long  as  the  sequence  contains  one  or 
more  integral  digits,  but  that  as  soon  as  the  sequence  is  a 
purely  decimal  number  the  mantissa  changes.  To  obviate 
this  difficulty  and  to  keep  the  mantissa  the  same  for  a  given 
sequence  of  digits  regardless  of  the  position  of  the  decimal 
point,  we  note  that  the  number  —  0.8911  may  be  written, 
without  change  of  value,  in  the  form  9.1089  —  10.  We 
have  added  10  and  subtracted  10,  and  have  therefore  left 
the  value  unchanged.  We  may  then  say 

log  .1285  =  -  0.8911  =  9.1089  -  10, 

and  if  we  agree  to  use  the  latter  formf  we  see  that  the 
mantissa  of  the  logarithm  of  .1285  (that  is,  1089)  is  the 
same  as  the  mantissa  of  the  logarithm  of  the  sequence  1285 
when  it  contains  integral  digits.  We  may  now  write 

log  1.285  =0.1089  log  .1285    =  9.1089  - 10 

log  12.85  =  1.1089  log  .01285  =  8.1089  -  10 


and  make  the  statement:  In  the  common  system  the  mantissa 
of  a  logarithm  is  unique  for  a  given  sequence  of  digits.     TJie 

*  For  example,  in  dividing  one  logarithm  by  another. 

t  This  form,  9.1089  —  10,  is  perfectly  convenient  as  long  as  the  opera- 
tions to  be  performed  are  addition  and  subtraction,  which  are  the  usual 
operations  in  dealing  with  logarithms. 

D 


34  PLANE   TRIGONOMETRY  [HI,  §31 

characteristic  is  one  less  than  the  number  of  integral  digits. 
If  a  number  be  purely  decimal,  count  the  decimal  point  and 
the  zeros  before  the  first  significant  figure.  The  result  sub- 
tracted from  10  minus  10  will  be  the  characteristic. 

32.    Four  Computation  Theorems.     The  use  of  logarithms 
in  computation  depends  upon  the  four  following  theorems  : 

I.  In  any  system  the  logarithm  of  a  product  is  equal  to  the 
sum  of  the  logarithms  of  its  factors. 

To  prove,  loga  mn  —  s  =  Iog0  m  -f  loga  n  +  •••  +  loga  s. 
Let  logd  m  =  x        then        ax  =  m 

loga  n  =  y  av  =  n 

loga  s  =z  a*  =  s. 

Whence      a*  •  av  —  a2  =  ax+v+ '"  +*  —  mn  —  s, 
or,  by  the  definition  of  a  logarithm, 

logtt  mn  -.  s  =  x  +  y  +  —  +  z. 

That  is,  loga  mn  —  s  =  Iog0  m  +  loga  n  +  —  +  loga  s. 

This  theorem  replaces  the  operation  of  multiplication  by 
the  simpler  operation  of  addition. 

II.  In  any  system  the  logarithm  of  a  quotient  is  equal  to 
the  logarithm  of  the   dividend  minus  the   logarithm  of  the 
divisor. 


To  prove, 

lOgfl  ~"  =  l°go  m  —  l°go  n" 

n 

Let 

loga  m  =  x 

then 

ax  =  m 

loga  7i  —  y 

ay  =  7i. 

Whence 

£«<*-» 

_m 

a" 

71 

Ill,  §32]     SOLUTION   OF   RIGHT   TRIANGLES  35 

or,  by  the  definition  of  a  logarithm, 


That  is, 

log«  —  =  log«  ra  -  logtt  n. 

tit 

This  theorem  replaces  the  operation  of  division  by  the 
simpler  operation  of  subtraction. 

III.  In  any  system  the  logarithm  of  a  power  of  a  number 
is  equal  to  the  exponent  of  the  power  times  the  logarithm  of  the 
number. 

To  prove,  loga  mn  =  n  Iog0  m. 

Let  loga  m  =  x  or  a*  =  m. 

Whence  (a*)n  =  anx  =  wT, 

or,  by  the  definition  of  a  logarithm, 

Iog0  mn  =  nx. 
That  is  logfl  w"  =  n  Iog0  m. 

This  theorem  replaces  the  operation  of.  involution,  or 
successive  multiplications,  by  the  simpler  operation  of  a 
single  multiplication. 

IV.  In  any  system  the  logarithm  of  a  root  of  a  number  is 
equal  to  the  quotient  of  the  logarithm  of  the  number  by  the 
index  of  the  root. 

To  prove  log(>V^  =  ^. 

n 

Let  logd  m  =  x  or  a*  =  m. 

Whence  ~\/ax  =  an  = 

or,  by  the  definition  of  a  logarithm, 


n 


36  PLANE    TRIGONOMETRY  [III,  §32 

That  is  loga^/m  =  1^^. 

n 

This  theorem  replaces  the  operation  of  evolution,  or 
extraction  of  roots,  by  the  simpler  operation  of  division. 

Another  theorem,  important  in  the  theory  of  logarithms, 
but  of  which  no  application  is  made  in  the  study  of 
trigonometry  is  the  following  : 


Proof  :     Let  logt  m  =  x     then     b1  =  ra 

Iog6  a  =  y  by  =  a. 

•L  -L 

Whence  m  —  bx  —  (by)  y  =  ay 

logam=* 

y 

By  means  of  this  theorem  the  logarithm  of  a  number  to 
any  base  can  be  found  if  the  logarithms  of  numbers  to  some 
one  base  are  known.  Thus,  assuming  that  logarithms  to 
the  base  10  are  known, 

W  71  21  =  lQg'»  71'24  -  '°g'°  7t24  -  1'8627  =  i  2650 
logloe        logw  2.718     0.4343 

As  a  corollary  of  the  above  theorem  we  have,  putting 
6  =  m,  1 

logam  =-  -- 
logma 

33.  Special  Properties  of  Logarithms.  In  addition  to  the 
preceding  theorems  we  may  note  the  following  properties 
of  logarithms  : 

1.  In  any  system  the  logarithm  of  1  is  0.     For,  by  the 
definition  of  zero  exponent,  a°  =  1  .     Therefore,  loga  1  =  0. 

2.  In  any  system  the  logarithm  of  the  base  is  1. 
For  a1  =  a.     Therefore,  loga  a  =  1. 


Ill,  §34]    SOLUTION   OP   RIGHT   TRIANGLES  37 

3.  In   any   system   whose   base   is   greater   than   1   the 

logarithm   of  0   is    —  oo.      For,   a  >  1,  or00  =  —  =  —  =  0. 

a00       °° 

Therefore,  loga  0  =  —  oo.  That  is,  the  base  of  the  system 
being  greater  than  unity,  the  logarithm  of  a  number  which 
becomes  smaller  than  any  assigned  number  however  small, 
is  negative  and  numerically  greater  than  any  assigned 
number  however  great. 

4.  The  cologarithm  of  a  number  is  the  logarithm  of  the 
reciprocal  of  the  number. 

Thus,  the  base  being  10, 

colog  n  =  log  -  =  log  1  —  log  n, 

n 
or, 

colog  n  =  0  —  log  w, 
which  may  be  written, 

colog  n  =  (10  —  10)  —  log  n. 

Therefore,  to  find   the  cologarithm  of  a  number  to   the 
base  10  subtract  the  logarithm  of  the  number  from  10  —  10. 
It  may  be  noted  that 

log  —  =  log  m  •  -  =  log  m  +  log  -  =  log  m  -f  colog  n. 
n  n  n 

Therefore  we  may,  instead  of  subtracting  the  logarithm  of 
a  number,  add  its  cologarithm.  It  is  found  convenient  to 
do  so  in  most  cases. 

34.    The    following   example   will  illustrate   the   use  of 
logarithms  in  making  numerical  computations. 

Example. 

Find  the  value  of 


.0005616  x  V-  424.65 
(6.73)4x  (.03194)* 

We  note  first  that,  with  the  definition  of  logarithms  we 
have  adopted,  negative  numbers  have  no  logarithms.     But 


38  PLANE   TRIGONOMETRY  [III,  §34 

the  numerical  result  of  operations  of  multiplication  and 
division  is  the  same  no  matter  what  the  combination  of 
algebraic  signs.  We  therefore  find  the  numerical  value  of 
any  expression,  treating  all  numbers  as  positive,  and  deter- 
mine the  algebraic  sign  of  the  result  by  considering  the 
operations  indicated.  Thus,  in  the  above  example  the 
factors  are  all  positive  except  -\/—  424.65.  Therefore,  the 
number  of  which  we  are  to  extract  the  cube  root  is  negative 
and  the  final  result  will  be  negative. 

log  .0005616  =  6.7494  -  10 
|  log  424.65  =  0.3754 
4  colog  6.73  =  6.6880  —  10 
|  colog  .03194  =  1.2464 

3)5.0592  -  10 
log  ^=8.3531 -10 

^=.02255. 
Therefore    

.0005616  x\/- 424.65 
(6.73)4  x  (.03194)* 

NOTE.  The  colog  6.73  =  9.1720  —  10,  which  being  mul- 
tiplied by  four  gives  36.6880  —  40 ;  subtracting  and  add- 
ing 30  this  becomes  6.6880  —  10,  the  desired  form  of  "  a 
number  minus  10."  Similarly  to  divide  5.0592  —  10  by 
three  we  first  add  and  subtract  20.  Also,  in  finding  five 
sixths  of  the  cologarithm  of  .03194,  we  first  multiply  by  5 
and  then  divide  by  6,  in  order  that  any  error  arising  from 
inexact  division  by  6  may  not  be  increased  5-fold. 

35.  We  may  now  return  to  the  problems  of  Art.  28  and 
solve  them  by  the  use  of  logarithms. 

Example  1.     Given  ^4  =  32°  16',   a  =  124,   (7=90°,  find 
Bj  b,  and  c. 
As  before, 

b  =  a  cot  Ay  c  = 


sin  A 


Ill,  §35]    SOLUTION  OF   RIGHT   TRIANGLES  39 

Therefore 

log  b  =  log  a  +  log  cot  A,        log  c  =  log  a  —  log  sin  A. 

log  a         =2.0934  log  a         =12.0934-10 

log  cot  A  =  0.1997  log  sin  A  =    9.7274  -  10 
log  b  =2.2931  logc=    2.3660 

b  =  196.4,  c  =  232.3. 

Example  2.     Given  a  =  50,  b  =  60,  C  =  90°. 
As  before, 

tan  A  =  - ,  therefore  log  tan  ^1  =  log  a  —  log  b. 
b 

log  a  =  11.6990  -  10 
log  6=    1.7782 


log  tan  A=    9.9208-10 

A  =  39°  48'. 
Also 

.  or  log  c  =  log  a  —  log  sin  A. 

log  a  =  11.6990 -10 
log  sin  A  =    9.8063  -  10 
logc=    1.8927 
c=78.1. 

It  must  be  emphasized  that  results  obtained  by  logarith- 
mic computation  are  approximate.  The  value  of  the  loga- 
rithm of  a  number  cannot,  in  general,  be  found  exactly,  but 
only  approximately  to  four,  five,  or  any  desired  number  of 
decimal  places.  TJie  results  of  numerical  computation  by 
means  of  logarithms  are  not,  in  any  case,  correct  beyond  the 
number  of  decimal  places  in  the  logarithms  used  to  make  the 
computation.  In  the  same  way,  the  values  of  the  trigono- 
metric functions  being,  in  general,  not  exact  but  approxi- 
mate to  four,  five,  or  more  decimal  places,  the  solutions  of 
triangles  got  by  their  use,  with  or  without  logarithms,  are 
approximate  solutions  only,  to  the  degree  of  accuracy  of  the 
tables  used. 


40  PLANE    TRIGONOMETRY  [III,  §35 

Indeed,  in  all  but  the  simplest  problems  in  applied 
mathematics  the  results  are  necessarily  approximate,  the 
data  of  a  problem  being  themselves  approximate.  It  is  use- 
less to  try  to  make  results  "  more  accurate  "  by  using  tables 
of  logarithms  or  other  functions  carried  to  seven  places 
when  the  data  are  correct  only  to,  say,  three  figures.  In 
general  if  data  are  given  to  three  figures,  three-place  tables 
should  be  used ;  if  to  seven  figures,  seven-place  tables,  etc. 
On  the  other  hand,  no  matter  to  how  many  figures  the  data 
may  be  given,  if  we  are  using,  say,  four-place  tables,  the  data 
should  be  used  and  results  found  to  four  figures  only.  To 
illustrate  these  points  the  following  simple  example  will  be 
worked  in  four  ways :  1.  by  actual  multiplication ;  2.  by 
using  four-place  tables  ;  3.  by  using  five-place  tables ;  4.  by 
using  seven-place  tables. 

Example.     Find  the  value  of 

123045  x  200368. 

1.  By  actual  multiplication  the  result  is  24,654,280,560. 

2.  log  123045 

123 1  0    =    5.0899 
log  200368 

200  |  4    =    5.3019 
log  product  =  10.3918 

product  =  24,650,000,000 

which  agrees  with  the  first  result  to  four  figures. 

3.  log  1230  1 45  =    5.09007 
log  2003  |  68=    5.30183 
log  product  =  10.39190 

product  =  24,655,000,000 

which  does  not  agree  with  the  first  result  to  the  fifth  figure. 
It  will  be  noted  that  there  was  an  accumulation  of  errors  all 
in  one  direction.  Result  3  is  nearer  to  result  1,  however, 
than  is  result  2. 


Ill,  §  35]     SOLUTION   OF   RIGHT   TRIANGLES  41 

4.    log  12304|  5=  5.0900640 
log  20036  8=  5.3018284 


log  product  =  10.3918924 

product  =  24,654,280,000 
which  agrees  with  the  first  result  to  seven  figures. 


EXAMPLES 

What  is  the  value  of 


1.     IQlog  7.218.  2.     loglO2-6994. 

3.  Given  log  2  =  0.3010,  log  3  =  0.4771,  find  log  12. 

4.  Prove  10lo?a+1  =  10  a. 

5.  Is  log  14  =  log  2.  log  7?     Why? 

6.  18^12  =  1??     Why? 

log  3.      3 


Find  the  value  of 

g     log  .00365  9      log  77.95 

log  .05312*  '    log  .00684* 

Find  the  value  of  x  in  the  following  equations  : 

10.  logics  =  3.  12.    logex  =  2. 

11.  Iogi0x=|.  13.   logex  =  f. 

14.  log™  x  .  loge  x  =  logio  «2. 

15.  a  logio  x  —  b  logio  x  =  a2  —  62. 

16.  loge  x3  -  loge  x2  =  5. 

17.  \  logio  z10  -  logio  z3  =  4. 

Express  as  the  logarithm  of  a  fraction  : 

18.  log  (x2  -  <z2)2  _  log  (x2  -  «2)  _  log  (x  +  a). 

19.  Iog\/x2  +  a2  -  log  v/x2  +  a2  +  log  (x2  +  a2)^. 
Solve  the  equations : 

20.  e*  +  e~x  =  2.  21.   e?  —  e~x  =  0. 

22.    e2^-1)  —  2  e1-1  +  1  =  0.  23.   e2'*-1)  +  2  ex~l  +  1=0. 

24.  Given  10*  =  400,  prove  that  x  =  2  +  logio  4. 

25.  Solve  the  equation 

a2e-<«  _  &2e-&*  —  o.  (Assume  a  >  6) 


42 


PLANE    TRIGONOMETRY 


[HI,  §35 


Find  the  value  of  ate-"*  —  bze~bx  : 

26.   When  x  =  — s —     — e_ .       (a  >  6) 


a  —  b 


27. 


a—  b 


(«>&) 


Compute  the  values  of  the  following  : 

28.   logel.  29.    log,  2.  30.   log,  3. 

By  means  of  logarithms  compute  the  values  of  : 


31.   Ioge4. 


32. 

S/.01236.            33.    VI.  1193. 
(56.333)^            36       99.02 

34.    V-.  002807. 
37     (-16.65)4 

38. 
41 

Vll.ll                       v'.  02983 

lQg771-2.             39.    [log  (.00915)  ]*. 
40.04 

62.85  x  V3111.59                    42     (56.3; 

\/-  .00986 
4Q        693.08 

log  .00598 
)3  x  \/56.3 

43. 
46. 

— 

999.9  x  .002008 

V.08888  x  40.19 
01)7                       45.    ( 

(.0002635)"^ 

.0001)~i 

V(.OOl)3.                      44.    (. 

J/ 

'485.7  x  22.01  x  11.79 

\ 

-  55.5  X  -  66.66 

(5362)  ~* 

47. 

V 

.00298  x  .00384 

4Q      (88)2 

X  (999)1 

v^632  x  .06302                                    v'lOOOOOO 

EXAMPLES:    SOLUTION  OF  RIGHT  TRIANGLES 

Solve  the  following  right  triangles,  and  find  their  areas  : 
1.    B  =  24°  23',              7.     6  =  .2072,                13.     6  =  156.6, 
6  =  .02126.                      a  =  .4212.                          c  =  856.4. 
2.    B  =  55°  45',              8.    A  =  82°  6',               14.    B  =  43°  46', 
c  =  4116.                          6  =  .08937.                       a  =  66650. 
3.    jB  =  43°30',               9.     a  =  .8478,                15.    B  =  74°  17', 
a  =  26185..                       c  =  1.234.                         b  =  .00002039. 
4.    a  =  77.38,                10.    B  =  60°  14',             16.   A  =  29°  56', 

c 

=  91.08. 

c  = 

.007745. 

c  = 

.0007814. 

5. 

B 

=  76°  34',             11. 

A  = 

14°  53', 

17.    6  = 

8.243, 

b 

=  2423. 

a  = 

1353. 

c  = 

9.275. 

6. 

A 

=  67°  47',             12. 

B  = 

39°  22', 

18.    B  = 

58°  39', 

c 

=  .00954. 

a  = 

121.2. 

c  = 

35.73. 

19.   A  =  35°  8', 

20. 

6  =  3814, 

a  =  17270. 

a  =  3651. 

Ill,  §35]    SOLUTION   OF   RIGHT   TRIANGLES  43 

PROBLEMS 

21.  A  road  rises  348.9  feet  in  a  horizontal  distance  of  one  half 
mile.    Another  road  rises  the  same  height  in  a  distance  of  3019  feet 
along  the  road.     Which  road  is  the  steeper  and  by  how  much  ? 

22.  From  a  ship  sailing  due  east  at  the  rate  of  7.6  miles  per  hour 
a  headland  bears  due  north  at  10.35  A.M.     At  12.46  P.M.  the  headland 
bears  33°  west  of  north.      How  far  was  the  headland  from  the  ship 
in  each  position  ? 

23.  At  a  distance  of  502.3  feet,  horizontally,  from  the  center  of  a 
bridge  the  sidewalk  rises  at  an  angle  of  elevation  of  5°.    The  roadway, 
beginning  203.5  feet  farther  away  from  the  center,  has  an  angle  of 
elevation  of  4°  25'.     If  a  pedestrian  and  a  team  enter  the  bridge  at 
the  same  moment,  which  will  reach  the  center  first,  the  man,  walking 
3.4  miles  per  hour,  or  the  team,  going  5.6  miles  per  hour  ? 

24.  A  flagpole  20  feet  long  stands  on  the  corner  of   a  building 
143.6  feet  high.     Find  the  angle  subtended  by  the  flagpole  from  a 
point  100  feet  distant  from  the  foot  of  the  building  in  a  horizontal 
line. 

25.  If  the  radius  of  a  circle  is  835.4  feet,  what  is  the  length  of  the 
chord  which  subtends  an  arc  of  45°  37'  ? 

26.  In  a  circle  whose  radius  is  35.37  inches  is  inscribed  a  regular 
polygon  of  fifteen  sides.     Find  the  length  of  a  side. 

27.  A  tree  214.8  feet  high  casts  a  shadow  167.4  feet  long.     How 
many  degrees  is  the  sun  above  the  horizon  ?      What  is  the  time  of 
day  if  the  sun  rose  at  six  o'clock  and  will  set  at  six  o'clock  ? 

[Assume  that  the  sun  passes  through  the  zenith.] 

28.  A  gable  roof  is  23.4  feet  high  and  90.6  feet  broad.     By  how 
much  must  the  height  be  reduced  to  reduce  the  pitch  of  the  roof  40 
per  cent  ? 

NOTE.     The  pitch  of  a  roof  is  the  angle  between  the  slope  of  the 
roof  and  the  horizontal  line. 

29.  From  the  top  of  a  cliff  378.6  feet  above  the  sea,  the  angles  of 
depression  of  a  boat  and  a  buoy,  in  line  with  the  observer,  are  found 
to  be  29°  20'  and  11°  50'  respectively.     Is  the  boat  or  the  buoy  farther 
from  the  base  of  the  cliff  ?     How  much  farther  ? 

30.  The  point  B  is  1249  feet  due  east  of  the  point  J.,  and  the  point 
C  is  376  feet  due  east  of  B.     The  angle  of  elevation  of  B  above  A  is 
9°  13'  ;  of  C  above  B,  7°  23' .    A  railroad  runs  from  A  to  C  via  B. 
What  is  the  increase  in  altitude  from  A  to  C  ? 


44  PLANE    TRIGONOMETRY  [III,  §35 

31.  If,  in  Example  30,  the  railway  could,  by  grading,  be  made  to 
run  in  a  straight  line  from  A  to  (7,  what  would  be  the  angle  of  eleva- 
tion of  the  new  route  ? 

32.  How  much  shorter  would  the  railway  of  Example  31  be  than 
the  railway  of  Example  30  ? 

33.  Taking  the  earth  as  a  sphere  of  radius  3956  miles,  what  is  the 
length  of  the  radius  of  the  Arctic  Circle,  latitude  66°  32'  N.  ? 

34.  Taking  the  earth  as  a  sphere  of  radius  3956  miles,  what  is  the 
latitude  of  a  place  which  is  2113  miles  from  the  earth's  axis  ? 

35.  A  vessel  sailing  due  south  at  a  uniform  rate  observes  at 
7.15  A.M.  that  a  lighthouse  bears  70°  east  of  south.     At  8.05  A.M.  the 
lighthouse  is  12.75  miles  due  east  from  the  ship.     How  far  from  the 
ship,  and  in  what  direction,  will  the  lighthouse  be  at  9.30  A.M.  ? 

36.  A  ship  sailing  due  south  at  a  uniform  rate  observes,  at  6  A.M., 
a  lighthouse  11.25  miles  away,  due  east.  At  6.30  A.M.  the  lighthouse 
bears  17°  57'  north  of  east.  What  will  be  the  bearings  of  the  light- 
house from  the  ship  at  9  A.M.  ?  How  fast  does  the  vessel  sail  ? 

37.  Taking  the  Earth  as  a  sphere  with  diameter  7912  miles,  what  is 
the  distance  of  the  farthest  point  on  the  Earth's  surface  visible  from 
the  top  of  a  mountain  8200  feet  in  height  ? 

38.  The  towns  B  and  C  lie  due  east  from  the  town  A,  B  being  half- 
way from  A  to  O,  which  are  5  miles  apart.      The  towns  5,  (7,  and  D 
are  equally  distant  from  each  other.     How  far  is  D  from  A  and  in 
what  direction  ? 

39.  A  ray  of  light  from  a  source,  A,  strikes  a  mirror,  102  mm. 
broad,  at  a  point  two  thirds  of  the  way  from  the  edge.     The  ray  is 
then  reflected  to  E  at  a  perpendicular  distance  25.7  mm.  from  the 
mirror.     Eind  the  length  of  the  path  traveled  by  the  ray. 

40.  From  a  window  of  a  house,  on  a  level  with  the  bottom  of  a 
spire,  the  angle  of  elevation  of  the  top  of  the  spire  was  41°.     From 
another  window,  20.5  feet  directly  above  the  former,  the  like  angle 
was  37°  31'.     What  was  the  height  of  the  spire  ? 

41.  Having  at  a  certain  (unknown)  distance  measured  the  angle  of 
elevation  of  a  cliff,  a  surveyor  walked  60  yards  on  a  level  toward  the 
cliff.     The  angle  of  elevation  from  this  second  station  was  the  com- 
plement of  the  former  angle.      The  surveyor  then  walked  20  yards 
nearer  the  cliff,  in  the  same  line,  and  found  the  angle  of   elevation 
from  the  third  station  to  be  double  the  first  angle.      How  high  was 
the  cliff  ? 


CHAPTER   IV 
FUNDAMENTAL  IDENTITIES 

36.  In  this  chapter  we  shall  discuss  some  of  the  impor- 
tant relations  of  analytical  trigonometry.     The  number  of 
such  relations  is,  of  course,  unlimited,  but  there  are  a  few, 
of   frequent   occurrence   and   of   fundamental   importance, 
upon  which  the  others  depend  ;  it  is  this  fundamental  group 
with  which  we  shall  now  deal.     Let  us  first  observe  how 
the  need  for  some  of  the  relations  may  arise.      We  have 
seen  (Art.  27)  that  as  the  angle  increases  from  0°  the  sine 
of  the  angle  also  increases.     But  does  the  sine  increase  at 
the  same  rate  as    the  angle,  so   that,  for  instance,  if   the 
angle  be  made  twice  as  large  the  sine  also  becomes  twice 
as  large  ?     This  is  obviously  not  so,  for,  as  we  have  seen, 
the  sine  of  60°  is  not  twice  the  sine  of  30°.     What  then 
are  the  relations,  if  there  be  any  such,  by  which  we  may 
find  the  functions  of  twice  an  angle  when  the  functions  of 
the  angle  are  given  ?   or  again,  is   there  any  relation  con- 
necting the  functions  of  the  sum  of  two  angles  with  the 
functions  of   the  angles  separately  ?      Such   questions   as 
these  we  shall  now  proceed  to  answer. 

37.  The  Addition   Formulae.     Let  x  and  y  be   two   acute 
angles,  whose   sum   may  be   an   angle   either   in   the  first 
quadrant  or  in  the  second.      Construct,  Fig.  18,  the  angle 
XOP  equal  to  x  and  add  to  it  the  angle  POQ  equal  to  y. 
Then  the  angle  XOQ  is  equal  to  x  +  y.     From  any  point, 
A)  in  the  terminal  side  of  the  combined  angle  x  4-  y  draw 
AB  perpendicular  to  the.  axis  of  x  which  is  the  initial  side 
of  the  angle  x.     Then  OB,  BA,  and  OA  are  respectively 
the  abscissa,  ordinate,  and  distance  of  the  point  A  and  we 

45 


46 


PLANE    TRIGONOMETRY 


[IV,  §37 


may  write  any  function  of  the  angle  x  +  y.  But  as  we 
wish  to  express  the  functions  of  x  +  y  in  terms  of  the  func- 
tions of  x  and  y,  we  proceed  to  draw  lines  which  will  give 
us  those  functions.  Thus,  from  A  draw  AC  perpendicular 
to  the  terminal  side  of  the  angle  x,  and  from  C  draw  CD 
perpendicular  to  the  axis  of  x  and  CE  perpendicular  to  AB. 


FIG.  18. 

Then  since  AE  is  perpendicular  to  OX  and  AC  to  OP,  the 
angle  EAC  is  equal  to  the  angle  x,  each  of  the  angles  being 
acute.  We  now  have 

,          .      BA     BE  +  EA     DC.EA 
w(x  +  y)=  —  =     --     =' 


But  these  last  two  ratios  are  not  functions  of  any  of  the 
angles  in  the  figure.  To  obtain  a  function  of  x  or  y  we 
must  use  with  DC  either  OD  or  OC,  and  with  OA  either 

DC 

OC  or  CA.     Therefore  we  shall  multiply  and  divide  - 

OA 

by  the  common  line  OC.  Similarly  with  EA  and  OA  we 
use  CA.  Thus  we  may  write 

v     DC     OC^EA      CA 

m(x+y)=—  +—    .- 


or 


sin  (x  +  y)  =  sin  x  cos  y  +  cos  x  sin  y. 


IV,  §38]  FUNDAMENTAL   IDENTITIES 

In  the  same  way 

N     OB      OD-BD     OD     EC 

^-— — -- 

OC_EC     AC* 
OC  '  OA     AC  '  OA9 


47 


or 


cos  (x  -f  y)  =  cos  x  cos  y  —  sin  x  sin  y. 


38.    The  Addition  Formulae  (continued).     Again  let  x  and  y 
be  two  acute  angles  where  x  may  be  either  greater  or  less 


FIG.  19. 

than  y.  Construct,  Fig.  19,  the  angle  XOP  equal  to  x  and 
from  it  subtract  the  angle  QOP  equal  to  y.  Then  the  angle 
XOQ  is  equal  to  x  —  y.  From  A,  any  point  in  the  termi- 
nal side  of  the  combined  angle,  draw  AB  perpendicular  to 
the  axis  of  x  which  is  the  initial  side  of  the  angle  x.  Then 
OB,  BA,  and  OA  are,  respectively,  the  abscissa,  ordinate 
and  distance  of  A.  From  A  draw  AC  perpendicular  to  the 
terminal  side  of  the  angle  x,  and  from  C  draw  CD  perpen- 
dicular to  the  axis  of  x  and  CE  perpendicular  to  BA  pro- 

we* 
*  Note  that  in  =£  we  use  AC  as  the  positive  direction  of  the  line, 


AC 


AC 


therefore  AC  must  be  positive  in  the  ratio  — ^  also. 


48  PLANE    TRIGONOMETRY  [IV,  §38 

duced.  Then  since  AE  is  perpendicular  to  OX  and  AC  to 
OP,  the  angle  EAC  is  equal  to  the  angle  x,  each  being 
acute.  We  now  have  as  in  Art.  37, 

.    ,          ,     BA      BE-AE     DC     AE 
**-= 


__ 
~  OC  '  OA     AC  '  OA\ 

or, 

sin  (x  —  y)  =  sin  x  cos  y  —  cos  x  sin  y. 


OA  OA  OAOA 


=  QD     OC     CM     CA 
~OC  '  OA     CA  '  OA' 

or, 

cos  (x  —  y)  =  cos  x  cos  y  -f  sin  x  sin  y. 

39.  We  have  thus  proved  the  formulae 

sin  (x  ±  y)  =  sin  x  cos  y  ±  cos  x  sin  y,  _ 

cos  (x  ±  y)  —  cos  x  cos  y  =F  sin  x  sin  y, 

for  values  of  x  and  ?/  less  than  90°.  It  now  remains  to  be 
proved  that  these  relations  are  true  for  all  values  of  x  and 
y.  This  may  be  done  by  a  geometric  construction  as  in  the 
cases  given,  but  the  following  method  is  preferable. 

40.  Let  x  be  an  angle  in  the  second  quadrant  and  y  an 
angle  in  the  third  quadrant.     Then  we  may  put  x  =  90°  +  a 
and  y  =  180°  +  &,  where  a  and  b  are  acute.     We  may  now 
write  _      _ 

cos  (x  +  ?/)  =  cos  (90°  -f  a  +  180°  +  6) 


=  cos  (270°  +  a  +  b) 

=  sin  (a  +  b).  (Art.  19) 

=  sin  a  cos  b  +  cos  a  sin  b.  (Art.  39) 

But  a  =  -  90°  +  x  and  6  =  -  180°  +  y. 


IV,  §40]          FUNDAMENTAL  IDENTITIES  49 

Therefore 


cos  (#  +  ?/)=  sin  (-  90°+#)  cos  (-  180°  +y)+  cos  (-  90°+z) 

sin(—  180°  +y) 

=  (  —  cos  x)  (  —  cos  y)  +  (sin  a;)  (  —  sin  y)      (Art.  19) 

or 

cos  (x  +  y)  =  cos  a;  cos  y  —  sin  #  sm  y 

which  is  the  same  as  the  relation  of  Art.  39. 

Again,  let  x  be  an  angle  in  the  first  quadrant  and  y  an 
angle  in  the  third.  We  may  put  y  =  180°  4-  6,  where  b  is 
acute,  and  write 


sin  (x  —  y)  =  sin  (x  —  180°  +  b) 


=  sin  (-180°  +  x  -b) 

=  -  sin  (x  -  b)  (Art.  19) 

=  —  sin  a;  cos  b  +  cos  x  sin  6.  (Art.  39) 

But  b      =  -  180°  +  y,  and  therefore, 

sin  (x  —  y)=  —  sin  x  cos(—  180°  +  y)  +  cos  x  sin  (—  180°  +y) 
—  —  sin  x(—  cos  y)  +  cos  x(—  sin  y)  (Art.  19) 

=  sin  x  cos  y  —  cos  x  sin  y. 

Thus  it  may  be  proved  that  the  equations  of  Art.  39  are 
true  for  all  values  of  x  and  y. 

The  importance  of  these  four  relations,  (10)  of  Art.  39, 
can  hardly  be  over-emphasized.  From,  them,  together  with 
those  given  in  Art.  12,  may  be  derived  all  other  trigono- 
metric identities.  The  method  of  so  doing  is  shown  in  the 
following  articles,  and  is  illustrated  by  the  following 
examples : 

Example  1.     Prove  the  relation 

sin  (45°  +  a)  cos  (45°  -  b)  +  cos  (45°  +  a)  sin  (45°  -  6) 

=  cos  (a  —  b). 


50  PLANE   TRIGONOMETRY  [IV,  §40 

This  is  simply  a  case  of  the  first  formula  of  (10)  where 

x  =  45°  +  a,  y  =  45°  -  b. 
We  may  write 

sin  (45°  +  a)  cos  (45°  -  6)  +  cos  (45°  +  a)  sin  (45°  -  b) 


=  sin  (45°  +  a  +  45°  —  b)=  sin  (90°  -fa  —  b)=  cos  (a  —  b). 

EXAMPLES 
Prove  that 

1.  sin  105°  +  cos  105°  =  cos  45°. 

2.  cos  (45°  -  x)  cos  (45°  +  x)  -  sin  (45°  -  x)  sin  (45°  +  x)  =  0. 

3.  sin  x  cos  (90°  -  x)  -  cos  x  sin  (90°  -  x)  =  -  cos  2  x. 

4.  cos  (30°  -  45°)  -  cos  (30°  +  45°)  =  sin  45°. 

5.  Given  sin  x  =  f ,  cos  y  =  f ,  find  sin  (x  +  y). 

6.  Given  cos  x  =  £,  cos  y  =  £,  find  cos  (x  —  y}. 

Given  tan  x  =  2,  tan  y  =  3,  find 

7.  sin  (x  +  y).     8.  cos  (x  +  y).     9.  sin  (x  -  y).     10.  cos  (x  -  y). 

41.  Tangent  of  a  Sum.  To  derive  an  expression  for  the 
tangent  of  the  sum  or  difference  of  two  angles  we  proceed 
as  follows : 


cos  (a;  ±  y) 

_  sin  x  cos  y  ±  cos  a;  sin  y 
""  cos  x  cos  y  T  sin  x  sin  y 

sin  x  cos  y  cos  #  sin  y 

__  cos  #  cos  y  cos  #  cos  y 

""  cos  a;  cos  y  sin  a;  sin  y 

cos  a;  cos  y  cos  a;  cos  y 


or  tan(*±y)  =  ^  ^«L.  (ii) 

1  =F  tan  JT  tan  y 

In  a  similar  manner  may  be  proved 
cot(x±y)  = 


coty±cota; 


IV,  §  42]          FUNDAMENTAL   IDENTITIES  51 

EXAMPLES 

Prove  that 


1  —  tan  x 

2.  tan  (45°  +  x)  tan  (135°  +  a)  +  1  =  0. 

3.  tan  (45°  +  x)  tan  (45°  —  x)  =  1. 

4.  Given  tan  a  =  2,  tan  6  =  4,  find  tan  (a  +  6). 

5.  Given  sin  a  =  £,  cos  6  =  $,  find  tan  (a  —  6). 

6.  Given  sec  a  =  3,  esc  6  =  4,  find  tan  (a  +  6). 

7.  Given  tan  a  =|,  tan  6  =  ^T,  find  a  +  6. 

8.  Given  sin  a  =  f  ,  cos  6  =  f  ,  find  tan  (a  +  6). 

9.  Given  sin  a  =  f  ,  sin  b  =  f  ,  find  a  +  6. 

Prove  the  following  identities. 

1Q     sin  (x  +  y)  _  cot  x  +  cot  y  ^ 
cos  (x  —  y)      1  +  cot  x  cot  y  " 

11.     tanx-tan(x-y)    = 
1  +  tan  x  tan  (x  —  ?/) 

42.  Functions   of   the  Double   Angle.    The   equations   of 
Arts.  39  and  41  being  true  for  all  values  of  x  and  y,  let  us 
assume  that  y  =  #.     Substituting  x  for  y  in  the  functions  of 
the  sum  of  two  angles  we  obtain 

sin  2  x  =  2  sin  x  cos  x  (13) 

cos  2  *  =  cos2  *  —  sin2  x.  (14) 

'*tan*   .  (15) 

1  —  tan2  x 

1.  (16) 

2  cot  a 

43.  The  student  should  clearly  understand  that  the  equa- 
tions of  Art.  42  give  the  values  of  functions  of  twice  an 
angle  in  terms  of  functions  of  the  angle,  no  matter  what 
the  value  or  form  of  the  angle  may  be.     For  example,  the 
following  relations  are  all  true,  being  merely  the  equations 
of  Art.  42  changed  slightly  in  form. 


52  PLANE    TRIGONOMETRY  [IV,  §  43 


2  sin  -  cos  -  ==  sin  a. 
2        2i 


cos        =  cos2       - sin2 
244 


2tan( 
tan  (2  a - 


-f) 


44.   Functions  of  the  Half-angle.     We  may  write  the  two 
proved  relations 

sin2  x  +  cos2  x  =  1  and  cos  2  a  =  cos2  x  —  sin2  a; 

in  the  form  «xt     .  2  x      1 

cos2  -  -|-  sm2  -  =  1. 


cos2  -  —  sin2  -  =  cos  x. 
2  2 


Subtracting  and  adding  these,  we  have 

os  jr. 

(17) 


2sin2?=l  -cos*. 


2  cos2    =  1  4-  cos  jr. 
2 

Dividing  the  last  two  equations   one  by  the   other   we 
obtain 


, 
2     1+cosJt 

(18) 


2     1  —  cos  x 

Thus  we  have  equations  which  give  the  sine,  cosine,  tangent, 
and  cotangent  of  one  half  an  angle  in  terms  of  the  cosine  of 
that  angle. 


IV,  §45]          FUNDAMENTAL   IDENTITIES  53 

EXAMPLES 

Given  sin  6  =  £,  find 

1.  sin  26.  3.   tan  26.  5.    cos-. 

2.  cos  26.  4.   sin  -.  6.   tan  |. 

2  2 

Prove  the  following  identities. 

7.  cos4  x  —  sin4  x  =  cos  2  x. 

8.  (sin  x  +  cos  x)2  =  1  +  sin  2  x. 

9.  tanx=     Sin2x     •  12.   tan  «  =  -*»*_. 

1  +  cos2x  2      1  +  cosx 

10. 


2       2secx  l+sin2x 


11.   tan     =     -.  14. 

2         sinx  2        secx 

15.    tan  (45°  +  a)  +  tan  (45°  -  x)  =  2  sec  2  x. 

45.   Sum  of  Sines  or  Cosines.     By  addition  and  subtraction 
of  the  two  equations 

sin  (x  +  y)  =  sin  x  cos  y  4-  cos  x  sin  y, 

sin  (x  —  y)  =  sin  a;  cos  y  —  cos  x  sin  y, 
we  obtain 

sin  (SB  +  y)  4-  sin  (a;  —  T/)  =  2  sin  a;  cos  y, 

sin  (SB  +  y)—  sin  (SB  —  y)=  2  cos  a;  sin  y. 

If  now  we  let  x  +-y  =  a,  SB  —  y  =  ft  so  that  cc  =  £(«  +  ft) 
and  y  =  ^(a  —  ft),  we  obtain  from  the  last  two  identities 

sin  a  +  sin  p  =  2  sin  £  (a  +  P)  cos|  (a  -  P), 
sina-sinp  =  2cos|(a+  P)sin|(a-  p). 

Proceeding  in  the  same  way  with  the  equations 

cos  (a;  ±  y)  =  cos  x  cos  y  T  sin  x  sin  y, 
we  obtain  two  more  equations  of  importance 

cosa  +  cos  p  =  2cos|(a  +  p)cos|(a  -  p), 

i  i  \    * 

cos  a  -  cos  p  =  -  2  sin  |  (a  +  p)  sin  \  (a  —  p). 


54  PLANE   TRIGONOMETRY  [IV,  §45 

EXAMPLES 

Express  each  of  the  following  as  the  algebraic  sum  of  sines  or 
cosines. 

1.  sin  6  x  cos  2  x.  4.    sin  (x  +  2  y)  cos  (x  —  y). 

2.  cos  4  x  sin  2  a;  5.   sin  (30°  +  x)  sin  (30°  -  x) . 

3.  cos-sin—.  6.    cos  3  x  cos  (x  —  y) . 

2          2 

Prove  the  following  identities. 

7.  cos  (30°  -  x)  cos  (60°  -  x)  =  £  (2  sin  2  x  +  V3). 

8.  cos  3  x  sin  2  x  —  cos  4  x  sin  x  =  cos  2  x  sin  x. 

9.  sin  x  cos  (x  4-  y)  —  cos  x  sin  (x  —  y)  =  cos  2  x  sin  y. 

46.  Identities  and  Equations.  It  should  be  borne  in  mind 
that  all  of  the  equations  of  this  chapter  are  identities,  that 
is,  they  are  true  no  matter  what  values  the  angles  may 
have.  We  shall  deal  later  on,  in  Chapter  VII,  with  trigo- 
nometric equations  of  condition,  where  we  shall  find  that 
not  every  value  but  only  particular  values  of  the  angles 
involved  will  satisfy  the  equations.  Also,  in  connection 
with  this  chapter  attention  should  be  again  called  to  the 
group  of  fundamental  identities  in  Art.  12. 

ILLUSTRATIVE  EXAMPLES 

Example  1.     Prove  that  sec  2  x  =  1  +  tan  x  tan  2  x. 

I 

=  -^—  = * =     cos's     =     sec*  a 

cos  2  x     cos2  x  —  sin2  x     1  —  tan2#     1  —  tan2a? 

_  1  +  tan2  x  _  ^        2  tan2  a; 
1  —  tan2  x  1  —  tan2  x 

=  1  +  tan  x  •     2tana;    =  1  +  tan  x  tan  2  x. 
1  —  tan2  x 

By  the  above  method  we  begin  with  sec  2  x  and  deduce 
or  derive  the  required  result.  Another  method  of  pro- 
cedure is  as  follows : 


IV,  §46]  FUNDAMENTAL   IDENTITIES  55 

Assume  that 


Then 


sec  2  x  =  1  +  tan  x  tan  2  x. 

=  l+H  2tan2a; 
1  —  tan2  x 

=  1  +  tan2  a; 
1  —  tan2  x 

_  cos2  x  4-  sin2  a; 
cos2  x  —  sin2  aj 


=  sec  2  #. 
Therefore,  the  original  assumption  is  correct.  ' 

Example  2.     Prove  that  esc  2  x  —  \  sec  x  esc  05. 
First  method. 

1  1  sec  x  esc  x 


esc  2  a;  = 


sin  2  a;     2  sin  x  cos  a; 


Second  method.     Take  the  reciprocals  of  both  members, 
sin  2  x  =  2  sin  x  cos  x. 

EXAMPLES 

Without  the  use  of  tables  find  the  following  : 

1.  Sine  and  cosine  of  15°.  3.   Tangent  of  15°. 

2.  Sine  and  cosine  of  22° 30'.  4.  Tangent  of  22°  30'. 

5.  Find  the  value  of  sin  3  x  in  terms  of  sin  x. 

6.  Find  the  value  of  cos  3  x  in  terms  of  cos  x. 

7.  Find  the  value  of  tan  3  x  in  terms  of  tan  x. 

8.  Find  the  value  of  tan  4  x  in  terms  of  tan  x. 

9.  Find  the  value  of  sin  4  x  in  terms  of  functions  of  x. 

10.  Find  the  value  of  cos  4  x  in  terms  of  functions  of  x. 

11.  Given  sin  4  x  =  a,  cos  4  x  =  6,  find  sin  8  x  and  cos  8  x. 

12.  Given  tan  3  x  =  a,  find  tan  6  x. 


56  PLANE    TRIGONOMETRY  [IV,  §46 

Prove  the  following  identities. 

13.  sin  (90°  -f  x  -f  y}  =  cos  x  cos  y  —  sin  x  sin  y. 

14.  sin  -  cos  -  +  cos  -  sin  -  =  sin  \  (a  +  6). 

22  22 

15.  cos  6  a  =  1  —  2  sin2  3  a. 

..       sin  4  a  -f  sin  2  a  _  3  —  tan2  a 


sin  4  a  —  sin  2  a     1  —  3  tan2  a 

17  1         _     sec20 

1  +  cos2  6  ~  tan2  e  +  2 ' 

18. 


19.  cos  5  x  cos  2  x  =  £  (cos  7  x  +  cos  3  x). 

20.  sec2x  esc2  x  =  sec2  x  +  esc2  x. 

21. 


sin  4x 
22.   cos  4  x  sin  x  =  \  (sin  5  x  —  sin  3  x)  . 

23 


sin3x  2 

24.   sinx= 


25.    sin  x  = 


1  -f  tan2  £  x 
2 


cot  £  x  +  tan  £  x 

26.    tan2x=  -  ?  -  . 
cot  x  —  tan  x 

27     8Jn(x  +  2y)-8in(x-. 
siny 

28.    cos  (2s-  y)  -  cos  (2x  +  y)  =  4  sin  x  sin  y< 
cosy 

29. 


30.   tan2    -  2  esc  x  tan-  -|-  1  =  0. 
2  2 


31. 


1  -f  tan2  x 

32.  tan  ^  x  =  esc  x  —  cot  x. 

33.  cot  -  =  esc  x  +  cot  x. 


IV,  §46]  FUNDAMENTAL   IDENTITIES  57 

2         1  +  sin  x  +  cos  x ' 

35'   2tan(x  +  45°)  = 

36.  Cosx-cos3x=:tan2x. 
sin  3  x  —  sin  x 

37.  sin  x  (1  +  tan  x)  +  cos  x  (1  +  cot  x)  =  sec  x  +  csc  x. 

38.  cot  x  —  cot  2  x  =  csc  2  x. 

og     1  +  sin  2  x  _  cos  x  +  sin  x 
cos2x         cos  x  — sin  x* 

40. 


_ 

41.  Given  sin  x  =  -  -      =  -  ,  find  the  value  of  cos  x. 

2.V-1 

42.  Given  r2  =  a2  sin  2  0, 

r2  cos  0  +  a2  cos  2  0  sin  0  _  tan3  ^ 
a2  cos  2  0  cos  6  —  r2  sin  0 

/i 

43.  Given  r  =  a  sec2  -  , 

Prove         r  cos  0  +  a  sec2  -  tan  -  sin  8 

99  /} 

-  2  -  2  -  =  _cot*. 
Oft  o 

a  sec2  -  tan  -  cos  6  —  r  sin  6 
2        2 

44.  Given  r  =  a  (1  —  cos  0), 

Prove  r  cos  0  -f  a  sin2  0  ,„  3  0 

-  —  uin  —  . 

a  sin  0  cos  6  —  rsmd  2 

Prove  the  following  identities. 

45.  cos3x  +  sm3x  =  2cot2Xt 

sin  x        cos  x 

46.  1  +  cos  2  x  cos  2  y  =  2  (sin2  x  sin  2  y  -f  cos  2  x  cos  2  y). 

47<  sin2  x  cos2  y-  cos2  x  sin2  y  =  ^       +       ^       _ 
cos2  x  cos2  y  —  sin2  x  sin2  y 

48. 


sn  x  —  cos  x 
49. 


sin  2  0  +  sin  d 


58  PLANE   TRIGONOMETRY  [IV,  §46 

50.    (sec  20  +  1)  Vsec2  0-1  =  tan  2  0. 

gl     cos  0  +  sin  0  __  cos  0  -  sin  0  _  2  tan  2  0. 
cos  0  —  sin  0     cos  6  +  sin  0 

52.    tan  2  0  -  sec  0  sin  0  =  tan  0  sec  2  0. 

53    1  —  cos  x  +  cos  y  —  cos  (x  +  y)  _  tan^a  ^ 
1  +  cos  x  —  cos  y  -  cos  (x  +  y)     tan  £  y ' 

54.  Given  x  =  —  3  cos2  0  sin  0  1      Prove  that 

y  =      3  sin2  0  cos  0}      2Vx2  +  y2  =  3  sin  2  0. 

55.  Given  x  =  a  cos  0  —  r  sin  0,     a  _  _   2  sin  2  0 

y  =  asin0  +  rcos0,  V2cos20' 

r2  =  2  cos  2  0,  Prove,  x2  +  ?/2  =  2  sec  2  0. 


CHAPTER   V 

THE  CIRCULAR  OR  RADIAN  MEASURE  OF  AN  ANGLE. 
INVERSE  TRIGONOMETRIC  FUNCTIONS 

47.  Circular  or  Radian  Measure  of  an  Angle.  Any  con- 
venient unit  may  be  chosen  for  the  measurement  of  angles. 
We  have  hitherto  used  the  degree,  subdivided  into  minutes 
and  seconds,  as  the  unit,*  but  we  shall  now  introduce 
another  unit  called  the  radian,  the  unit  angle  in  the  circular 
measure  of  angles. 

The  radian  is  an  angle  at  the  centre  of  a  circle  whose  sub- 
tending arc  is  equal  to  the  radius  of  the  circle. 


FIG.  20. 

It  is  obvious  that  the  radian  is  a  constant  angle,  is  the 
same  in  all  circles,  since  the  ratio  of  the  circumference  of  a 
circle  to  its  radius  is  constant. 

In  Fig.  20  let  the  angle  AOB  be  a  radian,  that  is,  let  the 
arc  AB  be  equal  to  OA,  the  radius  of  the  circle.  Also  let 
the  angle  AOC  be  an  angle  to  be  measured  in  radians.  To 
measure  a  quantity  is  to  find  its  ratio  to  another  quantity 
of  the  same  kind  chosen  as  the  unit.  Therefore, 

*  It  may  be  noted  that  the  right  angle  is  used  as  a  unit  in  the  study  of 
geometry,  partly  because  it  is  an  angle  easily  constructed. 

59 


60  PLANE   TRIGONOMETRY  [V,  §47 

Circular  measure  AOC  =  radians. 


AOC     arc  AC        arc  AC 


AOB     arc  AB     radius  OA 

Therefore,  to  measure  an  angle  in  circular  measure,  or  in 
other  words  to  express  the  angle  in  radians,  find  the  ratio 
of  the  arc  subtending  the  angle  in  any  circle  to  the  radius  of 
the  circle. 

If  we  represent  the  angle,  measured  in  radians,  by  x,  the 
length  of  the  arc  subtending  the  angle  by  s,  and  the  radius 

of  the  circle  by  r,  we  have  the  relation  x  =  - .     This  is  an 

algebraic  equation  involving  three  quantities.     If  any  two 
of  the  quantities  are  known  the  third  can  be  found.     Thus 

r  x 

Example  1.  What  is  the  radius  of  a  circle  in  which  an 
arc  of  12  inches  subtends  an  angle  of  \\  radians? 

r  =  -  =  — =8  inches. 
x     1£ 

Example  2.  If  the  radius  of  a  circle  is  15  feet  what 
length  of  arc  subtends  an  angle  of  two-thirds  of  a  radian  ? 

s  =  15  x  |=10  feet. 

We  know  that  the  ratio  of  a  semicircumference  to  its 
radius  is  IT  =  3.1416.  It  follows,  therefore,  that  the  angle 
which  is  sometimes  called  a  straight  angle,  and  which 
is  expressed  as  180°,  may  also  be  expressed  as  ?r  radians. 

Thus>  TT  radians  =  180° 

1  Gft° 

1  radian  =  —  =  57.296°,  approximately. 

7T 

Also  180°  =  TT  radians 

1°  =  -—  radians. 


V,  §47]    CIRCULAR  MEASURE   OF  AN   ANGLE          61 

By  means  of  these  two  relations  we  can  readily  reduce  a 
given  angle  from  either  system  of  measurement  to  the 
other. 

Example  1.     Express  the  angle  7  radians  *  in  degrees. 
TT  =  180°.     Therefore, 


Example  2.     Express  the  angle  —  -  in  degrees. 

o 

Since  TT  =  180°,          =     x  180°  =  120°. 


Example  3.     Express  the  angle  110°  32'  30"  in   circular 
measure.  110°  32'  30"  =  110£f°  =  ^|f  *°. 

Since        180°=7r,  therefore, 

10  _  _v__          2653°  _  jr_     2653  =  2653  TT 
"  180'  24     "180       24   ==  4320  " 

An  angle  in  circular  measure  is  usually  expressed  as  a 
multiple  of  IT. 

EXAMPLES 

Express  the  following  angles  in  degrees. 

1.   6.5  radians.  3.    -.  5.    1^. 

7  6 

o      2ir  /?      Sir 

*•    -g"  4.   3.8  radians.  6-    -g- 

Express  the  following  angles  in  circular  measure. 

7.  270°.  9.    25°  16'.  11.    208°  30'. 

8.  13°  24'.  10.    -450°.  12.    -98°. 
13.    What  is  the  ratio  of  a  radian  to  a  right  angle  ? 

*The  name  radian  is  often  omitted.    An  angle  written   w  and   read 

O 

"  pi  over  three,"  means  "  pi  over  three  "  radians,  or  about  1.05  radians. 


62  PLANE    TRIGONOMETRY  [V,  §  47 

How  many  right  angles  are  there  in  each  of  the  following  angles  ? 
14.    |.  15.    5^-.  16.    2*L.  17.    5  radians. 

18.  Through  how  many  radians  do  the  minute  and  hour  hands  of 
a  clock  turn  in  30  minutes  ? 

19.  Through  how  many  radians  does  the  minute  hand  of  a  clock 
turn  in  35  minutes  ? 

20.  Through  how  many  radians  does  the  hour  hand  of  a  clock  turn 
in  18  minutes  ? 

21.  The  front  wheel  of  a  cart  is  2  feet  in  diameter,  the  hind  wheel 
3  feet.     Through  how  many  radians  will  the  hind  wheel  turn  while 
the  front  wheel  is  turning  through  600°  ? 

22.  Through  how  many  radians  does  the  earth  revolve  about  its 
axis  in  a  week  ?     Is  the  result  the  same  in  45°  north  latitude  as  at  the 
equator  ? 

23.  A  wheel  turns  50  revolutions  per  minute.     Express  its  angular 
velocity  in  radians  per  second. 

24.  A  wheel  has  an  angular  velocity  of  20  radians  per  second. 
How  many  revolutions  does  it  make  per  minute  ? 

25.  Through  how  many  miles  will  a  point  on  the  equator  of  the 
earth  travel  as  the  earth  turns  through  1£  radians  ? 

26.  Through  how  many  miles  will  a  point  at  45°  north  latitude 
travel  as  the  earth  turns  through  one  radian  ? 

27.  The  radius  of  a  graduated  quadrant  is  2  feet,  and  the  grad- 
uations are  5'   apart.     What    is    the    distance  between   successive 
graduations  ? 

28.  What  must  be  the  radius  of  a  graduated  quadrant  if  the  dis- 
tance between  graduations  5'  apart  is  to  be  -fa  inch  ? 

48.  Inverse  Trigonometric  Functions.  Let  us  suppose  that 
y  is  the  sine  of  the  angle  x.  We  express  this  briefly  in 
mathematical  symbols  as  y  =  sin  x.  Suppose  now  that  we 
wish  to  make  the  inverse  statement  that  x  is  the  angle 
whose  sine  is  y.  To  express  this  in  mathematical  symbols 
we  write  x  =  sin"1  y,  where,  it  must  be  noted,  the  minus 
unity  is  not  an  exponent.  Having  expressed  our  idea  in 
symbols  we  next  note  that  x  depends  upon  y  for  its  value, 
is  a  function  of  y,  and  we  name  the  function  the  anti-sine  or 
inverse  sine.  Similarly  a  =  tan"1 6  means  that  a  is  the  angle 


V,  §49]     CIRCULAR  MEASURE    OF   AN   ANGLE 


63 


whose  tangent  is  6,  and  we  say  that  a  is  the  anti-tangent  of 
b.  In  this  way  we  have  a  group  of  six  inverse  trigonomet* 
ric  functions, 


sin-1  AT, 

5-1     V 


sec"1 x 


tan-1  AT, 

COS'*  AT,  COt^Af,  CSC"1  AT. 

These  six  quantities,  it  must  be  remembered,  are  angles. 


49.  General  Value  of  an  Angle.  Identities  connecting  the 
various  inverse  trigonometric  functions  exist  and  may  be 
derived  or  proved  by  methods  analogous  to  those  of  Chapter 
IV.  Before  taking  them  up,  however,  one  important  dif- 


FIG.  21. 


ference  between  the  direct  and  the  inverse  trigonometric 
functions  must  be  noted. 

If  y  =  sin  x,  and  if  we  give  x  a  particular  value,  say  30°, 
then  y  will  have  one  and  only  one  value,  one-half.  On  the 
other  hand,  if  x  =  sin"1  y  and  if  we  give  y  a  particular  value, 
say  -J-,  then  x  does  not  have  one  value  only  but  an  infinite 
number  of  values,  30°,  150°,  390°,  —  210°  etc.  This  being 
so  it  is  well  to  get  an  expression  that  will  represent  all  the 
angles  which  have  a  given  value  of  the  sine,  cosine,  etc. 

Let  sin  x  =  ±  a,  a  being  a  positive  number,  or  x  = 
sin~1(±  a)  and  let  a  be  the  smallest  angle*  which  has  for 

*  That  is,  a  if  we  use  positive  angles  only ;  a'  if  negative  angles  also 
are  used.  Either  method  may  be  adopted. 


64 


PLANE   TRIGONOMETRY 


[V,  §49 


its  sine  the  value  ±  a.     By  Fig.  21  we  see  that  the  possible 
values  of  x  are 

a,       TT  —  a,      2  TT  +  a,  3  TT  —  a,  4  TT  -f  a,  •  ••, 

—  TT  —  a,       —  2  TT  +  «,       —  3  TT  —  a,       —  4  TT  +  a,  •», 


which  may  be  written  in  the  general  form 
x—  siir1a  =  mr  +(  —  l)na 


(21) 


where  n  is  any  positive  or  negative  integer,  including  zero, 

and  a  is  the  least  angle  whose  sine  is  a.     This  is  called  the 

general  value  of  the  angle  and  a  is  called  the  principal  value. 

Since  the  cosecant  is  the  reciprocal  of  the  sine  we  may 


.  x          «  \ 

x  —  esc'1  a  =  mr  +  (-  l)na. 


/r\*~*\ 

(22) 


FIG.  22. 


Let  cos  x  =  ±  a  or  a;  =  cos-1(  ±  a)  and  let  «  be  the  least 
angle  whose  cosine  is  ±  a.     By  Fig.  22  we  see  that  possible 

values  of  x  are 

a,  2  TT  ±  a,  4  TT  ±  a,  •», 


or,  in  the  general  form 

x  =  cos"1  a  =  2  mr  ±  a 


(23) 


where  TI  is  any  positive  or  negative  integer,  including  zero, 
and  a  is  the  least  angle  whose  cosine  is  a. 

Since  sec  x  = we  may  write  for  the  general  value,  a 


cos  x 


V,  §49]     CIRCULAR  MEASURE   OF   AN   ANGLE  65 

being  the  principal  value, 

x  =  sec'1  a  =  2  TZTT  ±  a.  (24) 

Let  tan  x  =  ±  a  or  x  =  tan-1(  ±  a),  and  let  a  be  the  least 
angle  whose  tangent  is   ±  a.     By  Fig.  23  we  see  that  pos- 


FIG.  23. 
sible  values  of  x  are 

a,      TT  -f-  K)         2?r  +  «,         3  TT 

—  TT  +  a,     —  2  TT  +  a,     —  3  7T  +  a,  •  •  •, 

or,  for  the  general  value, 

JT  =  tan"1  a  =  mr  +  a 


(25) 


where  w  is  any  positive  or  negative  integer,  including  zero, 
and  a  is  the  principal  value. 

Giving  n  and  a  the  same  meaning  we  may  write,  since 
1 

(26) 


cot  # 


tana' 


x  =  cot"1  a  =  rnr  +  a. 


One  need  not  make  use  of  the  formulae,  21-26,  but  may 
proceed  as  follows  :  Find  the  two  smallest  angles,  positive  or 
negative,  which  correspond  to  the  given  value  of  the  func- 
tion. If  we  call  these  angles  a  and  /3  then  the  complete 
series  of  angles  will  be  given  by 

2  mr  +  a  and  2  mr  +  p. 


66  PLANE    TRIGONOMETRY  [V,  §49 

Example  1.     Write  the  general  value  of  cos'1 .9205. 
From  the  tables  of  trigonometric  functions  we  find  a  = 
23°.     Therefore 

23  7T 


x  =  cos-1 .9205  =  2  riTT  ±  23°  =  2 


180 


Example  2.     Write   the   general   value   of    sin'1  1.     We 
know  a  =  90°  =  -  •     Therefore 


a  =  sn'      =  ?nr       -». 

2i 

Example  3.     Prove  the  identity 

2  sin-1  a  =  sin'1  (2  a  Vl  -  a2). 

Let     sin"1  a  =  x,    then    sin  x  =  a.       Substituting     these 
values  in  the  formula  to  be  proved  we  have 


2  x  =  sin"1  (2  sin  #V1  —  sin2#), 
or 

sin  2  x  =  2  sin  a;  Vl  —  sin2  x  =  2  sin  x  cos  x. 

Q.  E.  D. 

Or,  we  may  proceed  as  follows  : 

We  know  sin  2  x  =  2  sin  x  cos  x,  which  may  be  written 

2  x  =  sin"1(2  sin  x  cos  x). 
Let  sin"1  a  =  x,  sin  x  =  a  and  substitute  : 

2  sin-1  a  ==  sin-1  (2  aVl  —  a2).  Q.  E.  D. 

Example  4.  Find  the  principal  value  of  tan-1 1  -f  tan-1  £. 
We  note  that  this  is  the  sum  of  two  angles  each  given  by 
the  value  of  its  tangent.  We  therefore  write,  formula  (11), 

tan  (tan-1  £  +  tan'1  £)  =  - 


Therefore,  tan-i  +  tan-ii  =  tan- 1  =  £ 

^  o  4 


V,  §49]    CIRCULAR  MEASURE   OF   AN   ANGLE          67 


EXAMPLES 

Write  the  general  values  of  the  following  angles  : 

1.  sin-i£.       3.  tan-!(-  1).      5.  Sec-i(--^rV     7.  sin-i(-—  Y 

V      \/3/  \       2  / 

2.  cos-iQ.     4.  cot-1-^-          6.  csc-!V2.  8.  cot-i(-VS). 

V3 

Find  the  value  of 

9.  sin  (sin-1  a).         11.   tan  (tan-1  y}.         13.   2  cos  (cos*1  .523). 
10.   cos-1  (cos  x).        12.   sec-1  (sec  30°).       14.    cot  (cot-1  2.718). 

15.  sin-1  (cos  35°).  /  2  \ 

18.    cos  f  tan-1  1  +  sec'1  —~  \  • 

16.  tan-1  (cot  40°).  ' 

19. 


17.   sin^sin-ii  +  cos-i-Ly        20.    coJtan'1  V3  +  tan-1— 

21.  Prove  that  x  =  sec-1  v/1  +  tan2  x. 

22.  Prove  that  tan-1  y  =  sec-1  vT+1/2. 

Prove  the  following  : 

23.  tan-i(  V2  +  1)  -  tan-1  (-  V2  —  1)  =  135°. 

24.  tan-1  V3  -  tan-1  f  -  —  \  =  tan"1  (-  3  V3). 

25.  tan-1  1  -  tan'1  (_  1)  =  tan'1  2. 

=  tan-1 
-  a 

b 


a2  -  2  62 
30.   sin-1  x  +  cos-1  y  =  tan- 


CHAPTER   VI 
THE   SOLUTION   OF   GENERAL   TRIANGLES 

50.  Four  Cases.  —  As  in  the  case  of  right  triangles  the 
solution  of  any  triangle  means  the  finding  of  the  values  of 
unknown  parts  from  the  parts  that  are  known.  Of  the  six 
parts  (three  angles  and  three  sides)  there  must  be  given 
three,  one  of  which  at  least  is  a  side,  in  order  that  the  tri- 
angle may  be  solved.  Consider  any  triangle,  Fig.  24. 


FIG.  24. 

The  following  cases  may  be  enumerated : 

I.  Given  a  side  and  two  angles;  say,  a,  A,  B. 

II.  Given  two  sides  and  the  angle  opposite  one  of  them  ; 
say,  a,  b,  A. 

III.  Given  two  sides  and  the  included  angle;  say,  a,  b,  C. 

IV.  Given  the  three  sides;  a,  b,  c. 

51.   The  Law  of  Sines.  —  Cases  I  and  II  may  be  solved  by 
means  of  the  following  theorem. 

In  any  triangle  the  sides  are  proportional  to  the  sines  of  the 
opposite  angles.     That  is,  Fig.  25, 

a :  b :  c  =  sin  A  :  sin  B :  sin  C.  (27) 

68 


VI,  §52]     SOLUTION   OF   GENERAL   TRIANGLES        69 
c  c 


A  D 

FIG.  25. 


Proof :  In  the  triangle  BAC  draw  CD  perpendicular  to 
BA.     Then 


Therefore 


AC 


BC      a 


DC  =  b  sin  A  —  a  sin  B. 


Whence 


a  :  b  =  sin  A  :  sin  B. 

Similarly  the  theorem  may  be  proved  for  the  other  pairs 
of  sides  and  angles. 

52.   Case  I.     Given  a  side  and  two  angles  ;  a,  A,  C. 
To  find  the  third  angle  we  have 

To  find  B  and  C  we  have 

&  _  sin  B         c___  sin  C 
a     sin  A '       a     sin  A ' 

selecting  in  each  case  that  proportion,  from  (27),  which  in- 
volves an  unknown  side,  b  or  c,  and  three  known  parts. 

From  these  two  proportions  we  have 

log  b  =  log  a  4-  log  sin  B  +  colog  sin  A 
log  c  =  log  a  +  log  sin  C  +  colog  sin  A 


70  PLANE   TRIGONOMETRY  [VI,  §52 

Example.     Given  a  =  412.7,  A  =  50°  38',  C  =  69°  13',  find 
B*  bj°'  B  =  180°  - 119°  51'  =  60°  9'. 

log  a  =  2.6157  log  a  =  2.6157 

log  sin  B  =  9.9382  log  sin  C  =  9.9708 

colog  sin  A  =  0.1118  colog  sin  A  =  0.1118 

log  b  =  2.6657  log  c  =  2.6983 

b  =  463.1  c  =  499.2 

53.   Case  II.     Given  two  sides  and  the  angle  opposite  one  of 
them;  a,  6,  A. 

We  have,  to  find  B, 

sin  B     b  t 

sinJ.     a 

Whence,  log  sin  B  =  log  sin  ^4  +  log  b  +  colog  a. 
Also,  (7=180°-  (^  +  B). 

Then  '=«*£ 

a      sin  .4 

Whence          log  c  =  log  a  +  log  sin  C  -f-  colog  sin  A. 
Example.     Given   a  =  31.24,    b  =  49,    A  =  32°  18',  find 

*'  ^C'  log  sin  ^1  =  9.7278 

log  6  =  1.6902 

colog  a  =  8.5053 

log  sin  B  =  9.9233 

B  =  56°  56' 

But  since  B  is  found  from  the  log  sine  it  may  have  two 
values  ;  namely,  56°  56'  and  180°  -  56°  56'  =  123°  4'.  To 
determine  which  value  is  correct  or  whether  both  are 
possible  we  recall  the  theorem  of  geometry  which  states  that 
if  the  given  angle  is  acute  and  the  side  opposite  is  less  than 
the  other  given  side,  then  it  may  be  possible  to  construct 
two  triangles  from  the  given  parts,  two  sides  and  an  oppo- 


VI,  §54]     SOLUTION   OF   GENERAL   TRIANGLES        71 


site  angle.  In  the  above  example  the  given  angle  A  is  acute 
and  its  opposite  side  a  is  less  than  6  ;  there  are  two  solutions 
and  both  values  of  B  must  be  used.  Figure  26  explains  the 
case  graphically. 


31.24 


=  32M8' 


Bi=56°56'  A =32°  18' 

FIG.  26. 


b=49 


a=31.24 


B2) 


Consequently  there  are  two  values  of  (7,  namely, 

d  =  180°  -  (A  +  B,)  <72  =  180°  -  (A 

=  90°  46'  =  24°  38' 

and  two  values  of  c,  got  as  follows : 

log  a  =  1.4947  log  a  =  1.4947 

log  sin  d  =  0.0000  log  sin  C2  =  9.6199 

colog  sin  A  =  0.2722  colog  sin  A  =  0.2722 

log  d  =  1.7669  log  c2  =  1.3868 

cx  =  58.46  c2  =  24.37 

If  the  given  angle  be  obtuse  there  will  be  only  one  solu- 
tion. If  the  given  angle,  A,  be  acute  and  the  side  a  be 
greater  than  the  side  6,  there  will  be  one  solution  only.  If 
A  be  acute  and  a  be  equal  to  the  perpendicular  from  C  to 
ABj  there  will  be  only  one  solution,  a  right  triangle.  In 
this  case  B  =  90°  and  log  sin  B  =  0.0000.  If,  A  being  acute, 
a  be  less  than  the  perpendicular  from  C  to  AB,  there  is  no 
solution.  In  this  case  log  sin  B  will  be  greater  than  zero, 
which  is  impossible  since  sin  B  cannot  be  greater  than  unity. 

54.  Case  III  may  be  solved  by  means  of  the  theorem 
following : 

In  any  triangle  the  sum  of  two  sides  is  to  their  difference  as 
the  tangent  of  half  the  sum  of  the  angles  opposite  the  two  sides 
is  to  the  tangent  of  half  their  difference. 


72  PLANE    TRIGONOMETRY  [VI,  §54 

Proof  :  By  Art.  51 

a  :  b  =  sin  A  :  sin  B. 
Whence,  by  composition  and  division, 

a  4-  6  __  sin  A  +  sin  B  __  sin  j-  (A  +  jB)  cos  |(  J.  —  ff) 
a  —  b     sin  .4  —  sin  B     cos  £(^L  +  5)  sin  \(A  —  B) 
or, 

a  +  b      tan      ^i  +  £)  ~ 


55.   Case  III.*     Given  two  sides  and  the  included  angle; 
b,c,A. 

By  Art.  54  we  have 

C       &  —  c 


tan 
The  sides  &  and  c  are  known,  and  also 

l(B  4-  C)=  |(1800  -  ^)=  90°  -  £ 
since  ^4  +  5  +  C  =  180°. 

Therefore  we  may  write 

tan  i(J5  -  (7)=  ^^  -  tan  ^(S  +  C), 


or, 

log  tan  ^(jB-<7)=log(&-c)  +  colog(&+c)+  log  tan 


Thus  -|  (5  —  C)  is  found,  and  by  finding  the  sum  and  dif- 
ference of  %(B  +  0)  and  *(B  —  C)  the  values  of  B  and  O 
are  known.  Finally,  to  determine  a  we  have,  as  in  Case  I  : 

a  :  b  =  sin  A  :  sin  B. 

Example.  Given  6  =  .06239,  c  =  .02348,  A  =  110°  32'  ; 
find  J5,  C,  a. 

+  O)=  90°  -  J  A  =  90°  -  55°  16'  =  34°  44'. 
&  +  c  =  .08587  6  -  c  =  .03891. 

*  See  also  Art.  61  following. 


VI,  §56]     SOLUTION    OF   GENERAL   TRIANGLES        73 


Whence  we  have 

log  (b  -c)=  8.5900 
colog(6  +  c)  =  1.0661 
logtanfr(JB+  C)  =9.8409 
log  tan  KJB  -  (7)=  9.4970 
-  0)=17°26'. 

+  C)=34°44' 
C  =  17°  18'. 


And  as 

We  have     B  =  52°  10' 


Then  log  b  =  8.7951 

log  sin  A  =  9.9715 

colog  sin  B  =  0.1025 

log  a  =  8.8691 

a  =  .07398 

56.  The  Law  of  Cosines.  Case  IV  may  be  solved  by  means 
of  the  following  theorem  : 

In  a  triangle  the  square  of  any  side  is  equal  to  the  sum  of 
the  squares  of  the  other  two  sides  minus  twice  the  product  of 
those  sides  by  the  cosine  of  their  included  angle. 

c 


That  is,  Fig.  27,  where  CD  is  perpendicular  to  AB, 
a2  =  b2  4-  c2  —  2  be  cos  A. 

We  have  proved  the  geometrical  theorem 
a2  =  bz  +  c2 -  2 c  •  AD* 


But 


cos  A  = ,  or  AD  =  b  cos  A. 

b 


*  Note  that  in  the  first  two  triangles  of  Fig.  27,  AD,  the  projection  of  6, 
is  read  left  to  right  and  is  positive ;  in  the  third  triangle  from  right  to  left 
and  is  negative. 


74  PLANE    TRIGONOMETRY  [VI,  §  56 

Therefore,  a2  =  b2  +  c2  -  2  6c  cos  A  Q.E.D. 

Obviously 


2  be 


(29) 


and  in  the  same  way 


2  ca  2  a& 

so  that  the  three  angles  may  be  found. 

57.  The  objection  to  the  formulae  of  Art.  56  is  that  they 
are  not  adapted  to  logarithmic  computation.  To  remove 
this  objection  we  proceed  as  follows  :  From  (29)  we  have 


-  cos  .   =  1 


2bc  2bc  2  be 


2  be 
or,  by  (17),  Art.  44, 


2  be 

Let  a  +  b  +  c  =  2  s. 

Then  a  -  6  +  c  =2(s  -  b),  a  +  b  -  c  =  2(s-  c), 

Whence  .     x    .  _  2(s  -  6)  •  2(*  -  c) 


--  (30) 

oc 
Similarly 


ca  ^  ab 

These  formulas  may  be  used  for  logarithmic  computation. 


VI,  §59]     SOLUTION   OF   GENERAL  TRIANGLES        75 
58.   Again,  from  (29),  Art.  56,  we  have 


c)(b  +  c—  a) 


2  be 


or,  by  (17),  Art.  44, 
o 


c-a) 


26c 


As  before,  letting  a 


=  2s,  this  becomes 


or 


cos 


Similarly 


s(s  —  a) 

^-v^H- 


(31) 


cos  i  (7  = 


59.  These  formulae  also  may  be  used  for  logarithmic 
computation,  but  a  more  convenient  set  is  obtained  by 
dividing  the  formulae  of  Art.  57  by  those  of  Art.  58.  We 
thus  obtain  

//o  fcWo  W\ 

(32) 


J 


A  comparison  of  these  three  sets  of  formulae  (30),  (31), 
(32),  will  show  that  for  the  complete  solution  of  a  triangle 
when  the  three  sides  are  given,  the  first  set  (30),  requires 
six  different  logarithms,  the  second  set  seven,  the  third  set 
four.  In  addition  to  this  slight  advantage  the  tangent  set, 
(32),  gives  more  accurate  results  than  the  other  two  when 
the  angles  involved  happen  to  be  very  small  or  very  near 
ninety  degrees. 


76  PLANE   TRIGONOMETRY  [VI,  §  60 

60.   Case  IV.      We  will  now  solve  a  triangle  when  the 
three  sides  are  known. 

Example.     Given  a  =  10,  b  =  12,  c  =  14  ;  find  A,  B,  C. 
Here  2  s  =  a  +  6  +  c  =  36,  so  that  we  have 

5  =  18,  log  s  =  1.2553,  colog  s  =  8.7447  -  10. 

s  -  a  =  8,  log  (s  -  a)  =  0.9031,  colog  (s  -  a)  =  9.0969  - 10. 
s  _  b  =  6,  log  (s  -  b)  =  0.7782,  colog  (s  -  b)  =  9.2218  -  10. 
s  -  c  =  4,  log  («  -  c)  =  0.6021,  colog  (s  -  c)  =  9.3979  -  10. 

log  (s-b)=    0.7782  log  (s  -  c)  =    0.6021 

log  (s  -  c)  =    0.6021  log  (s  -  a)  =    0.9031 

colog  s  =    8.7447  colog  s  =    8.7447 

colog  Q- a)  =    9.0969  colog  (s-  b)  =    9.2218 

2)19.2219  2)19.4717 


.log  tan  |  A  =  9.6110,  log  tan  ±B  =  9.7359, 

|-4  =  22°  13',  |  £  =  28°  34', 

^4  =  44°  26',  J5=57°    8', 

log  («- a)  =    0.9031 

log  (s  -  6)  =    0.7782 

colog  ,s=    8.7447 

cologO-c)=   9.3979 

219.8239 


log  tan i(7=9.9120, 

2  C=  78°  28'! 
CHECK:  A  +B  +  C=  180°  2'. 

A  common  method  of  solving  Case  IV  is  by  means  of  an 
auxiliary  quantity, 

r".v^    ^ 

We  may  write 

i  r  i  ,,         r  \  „        * 

\.&n(tA  = ,        tan -5= -,         tan-C  =  —  —  • 

s—  a  s—  b  s—  c 


VI,  §61]     SOLUTION    OF   GENERAL   TRIANGLES        77 

In  using  this  method  log  r  is  first  found,  whence  the  log- 
tangents  of  the  three  half-angles  are  readily  obtained. 

61.  Case  III ;  Other  Methods  of  Solution.  The  formulae 
of  Art.  56  may  sometimes  be  used  to  advantage  in  solving 
Case  III. 

Example  (see  Art.  55).  Given  b  =  .06239,  c  =  .02348, 
A  =  110°  32',  to  solve  the  triangle. 

We  have  a2  =  62  +  c2  —  2  be  cos  A. 

Then 

a2  =  (.06239)2  +  (.02348)2  -  2(.06239)  (.02348)  cos  110°  32'. 
log  b  =  8.7951  log  c  =  8.3708 

2  2 

log  b2  =  7.5902  log  c2  =  6.7416 

62  =  .003893  c2=    .0005516 

+  c2=  .0005516  log  2  =  0.3010 

-26ccos^=    .001028  log  b  =  8.7951 

a2=    .005473  log  c  =  8.3708 

log  cos  A  =  9.5450 
log  2  be  cos  A  =  7.0119 
2  be  cos  A  =  —  .001028 

Then     log  a  =  $  log  a2  =  8.8691  and  a  =  .07398. 

To  find  B  and  C  we  have  the  formulae  of  Art.  51. 
The  above  computation  can  in  some  cases  be  done  best 
and  quickest  without  the  use  of  logarithms. 

Another   Method  of   Solution   for   Case   III,   preferred   by 

many,  is  as  follows  : 

From  Fig.  27  we  see  that 

DC  =  b  sin  A,  AD  =  b  cos  A,DB  =  c-  AD. 


78 


PLANE   TRIGONOMETRY 


[VI,  §61 


nc1 

Then  tan  J3  =  —  ,  whence  B  is  known,  and  (7=180°  — 


(A  +  B).     To  find  a  we  use  a  =  - 


sin 


Applying  this  method  to  the  example  above  we  have 


log  b  =  8.7951 

log  sin  A  =  9.9715 

log  7)0=8.7666 

log  7)^  =  8.6567 

log  tan  B  =  0.1099 

J3=52°10' 


log  b  =  8.7951 
log-cos  A  =  9.5450 
log--4Z>  =  8.3401 

AD  =  -  .02188 
DB  ='.02348  +  .02188  =  .04536. 

log  DC  =  8.7666 
C  =  17°18'  log  sin  B  =  9.8975 

log  a  =  8.8691 
a=    .07398 

NOTE.  The  fundamental  importance  of  the  law  'of  sines  and  the 
law  of  cosines  should  be  noted.  By  their  use,  direct  or  indirect,  any 
triangle  whatever  may  be  solved. 

AREAS   OF  TRIANGLES 
62.   Right  Triangles. 

Case  I.    Given  the  two  legs  a  and  6,  Fig.  28. 

B 


FIG.  28. 

Representing  the  area  of  the  triangle  by  K,  it  is  obvious 
2K=ab.  (33) 


Case  II.     Given    the    hypotenuse  and    an  acute    angle,   c 
and  A. 


VI,  §63]     SOLUTION   OP   GENERAL   TRIANGLES        79 
Then  a  =  c  sin  A,     b  =  c  cos  A. 

Whence,  by  (33), 


2  K=  c2  sin  A  cos  ^4  =  £  c2  sin 
or  4  jRT  =  c2  sin  2  -4. 

Case  III.     Given  an  angle  and  the  adjacent  leg,  A  and  b. 
Then  a  =  b  tan  A. 

Whence,  by  (33), 

2K=  W  tan  A. 

Case  IV.     Given  the  hypotenuse  and  a  leg,  c  and  a. 
Then  62  =  c2  —  a2  or  b  =  V(c  +  a)(c  —  a). 

Whence,  by  (33), 


63.   Oblique  Triangles. 

Case  I.     Given  two  sides  and  the  included  angle,  a,'  6,  (7. 


In  Fig.  29  the  line  ^4.D  is  perpendicular  to  BC. 
It  is  obvious  that  2  7T=  a  X 


But 
Therefore, 


sin  C=        ,  or  ZL4  =  6  sin  C. 

b 


(34) 


80  PLANE   TRIGONOMETRY  [VI,  §63 

Case  II.     Given  a  side  and  the  angles,  a.  A,  B,  C. 
By  Art.  51, 

6  _  smB         ^  __  a  sin  B 
a      sin  A  sin  A 

Whence,  by  (34), 

2  K  =  fl2  8*n  -B  •  sin  C  ,ggx 

sin  .4 

Case  III.     Given  the  three  sides,  a,  &,  c. 
The  formula  (34),  of  Case  I  may  be  written,  by  Art.  42, 
(13)'  2  K=  2  ab  sin  -t  C  .  cos  £<7. 

Substituting  in  this  the  values  of  sin  -J  (7  and  cos  \  C  given 
in  Arts.  57  and  58,  we  have 


2  A'  =  2 


or  JT  =  Vs(s  -  d)(s  -  b)(s  -  c).  (36) 

EXAMPLES 

Solve  the  following  triangles,  in  each  case  obtaining  also  the  area 
of  the  triangle  : 


1. 

a 
B 
C 

= 

1419, 
29°  59', 
16°  1'. 

6. 

c 
A 
C 

= 

5141, 
96°  3', 
55°  46'. 

11.  6 
a 
B 

= 

.5042, 
.3618, 
74°  43'. 

2. 

a 

b 
B 

— 

3.384, 
9.828, 
109°. 

7. 

b 
c 
A 

= 

56.2, 
63.9, 
71°  33'. 

12.  a 

b 
c 

= 

.03574, 
.02921, 
.01853. 

3. 

a 
b 
C 

= 

302, 
427, 
134°  29'. 

8. 

b 
c 
C 

= 

268.5, 
282.9, 
75°  20'. 

13.  6 
a 
A 

= 

.2792, 
.2271, 
65°  45'. 

4. 

a 
b 
c 

= 

56.22, 
63.91, 
70.54. 

9. 

b 
A 
C 

=  6.362, 
=  76°  13', 
=  35°  17'. 

14.  a 
A 
B 

= 

.01044, 

26°  32', 
146°  26'. 

5. 

b 
c 
B 

= 

38.65, 
48.12, 
34°  32'. 

10. 

a 

c 
A 

= 

5499, 
2959, 
133°  34'. 

15.  a 
b 
B 

= 

31.49, 
49.88, 
44°  35'. 

VI,  §63]     SOLUTION   OP   GENERAL   TRIANGLES        81 

16.  c  =  .0357,      18.  b  =  4621,       20.  a  =  6.743, 
a  =  .0292,          a  =  6473,          b  =  3.025, 

5  =  31°  7'.         B  =  31°  7'.          c  =  4.271. 

17.  a  =  32.15,      19.  6  =  .4312,      21.  c  =  .01825, 
6  =  67.54,          c  =  .8901,          6  =  .02893, 

A  =  28°  26'.         A  -  29°  55'.         B  =  83°  30'. 

PROBLEMS  IN  THE  SOLUTION  OF  TRIANGLES 

22.  A  man  owns  a  triangular  lot  on  the  corner  of  two  streets 
which  do  not  intersect  at  right  angles.    The  frontage-  on  one  street  is 
300  feet,  on  the  other  250  feet.     The  back  line  of  the  lot  is  350  feet 
long.     If  he  buys  land  to  add  275  feet  to  the  300-foot  frontage,  by  how 
much  is  his  lot  increased  in  size  ? 

23.  A  man  owns  a  triangular  lot  on  the  corner  of  two  streets 
which  intersect  at  an  angle  of  62°.     The  frontage  on  one  street  is 
200  feet,  on  the  other  150  feet.     If  the  land  is  worth  one  dollar  a 
square  foot  and  the  man  has  $  1200  with  which  to  increase  the  size  of 
his  lot,  by  how  much  can  he  lengthen  the  150-foot  frontage  ? 

24.  The  perimeter  of  a  triangle  is  100  feet,  and  the  perpendicular 
from  the  vertex  C  to  the  base  AB  is  30  feet.     The  angle  A  is  50°. 
Find  the  length  of  the  base  AB. 

25 .  What  is  the  perpendicular  height  of  a  hill  which  is  known  to  rise 
72  feet  for  every  100  feet  of  length  of  its  slope,  if  the  angle  of  elevation  of 
the  hilltop  from  a  point  100  yards  from  the  base  of  the  hill  is  31°  ? 

26.  From  where  I  stand,  50  feet  from  the  bank  of  a  stream,  the 
angles  of  depression  of  the  near  and  far  banks  of  the  stream   are 
respectively  15°  37'  and  6°  24'.     How  wide  is  the  stream  ?     How  far 
am  I  above  the  level  of  the  stream  ? 

27.  A  man  5  feet  6  inches  tall,  standing  on  a  bluff  40  feet  high, 
measures  the  angles  of  depression  of  the  near  and  far  shores  of  a  bay. 
The  angles  are  46°  52'  and  5°  3'  respectively.     How  wide  is  the  bay  ? 

28.  A  man  5  feet  tall,  standing  on  the  edge  of  a  pond,  finds  the 
angle  of  elevation  of  the  top  of  a  tree  on  the  other  bank  to  be  44°  26'. 
The  angle  of  depression  of  the  reflection  of  the  treetop  is  60°  47'.    Find 
the  height  of  the  tree. 

The  reflection  of  an  object  appears  as  far  below  the  surface  as  the 
object  is  above  the  surface. 

29.  The  frontage  on  the  beach  (AB)  of  a  quadrangular  lot  ABCD 
cannot  be  measured.     The  sides  BC,  CD,  and  DA  are  found  to  be 
236,  155  and  105  feet  respectively.     The  angles  DJ.Cand  DBG  are 
32°  20'  and  29°  50'  respectively.     Find  the  length  of  AB. 

G 


82  PLANE    TRIGONOMETRY  [VI,  §  63 

30.  The  bases  of  a  trapezoid  are  48.25  and  94.75  feet.     The  angles 
at^the  ends  of  the  longer  base  are  63°  52'  and  70°  55'.     Find^the  lengths 
of  the  other  two  sides. 

31.  Two  sides  of  a  triangle  are  8.53  and  7.41.     The  difference 
between  the  angles  opposite  these  sides  is  18°  23'.     Solve  the  triangle. 

32.  The  area  of  a  triangle  is  979  square  feet.     The  angle  A  is 
56°  22'  and  the  side  b  is  44.80  feet.     Solve  the  triangle. 

33.  Two  sides  of  a  parallelogram  are  8005  and  5008.     The  included 
angle  is  60°  53'.    Find  the  lengths  of  the  diagonals. 

34.  The  diagonals  of  a  quadrangular  field  ABCD  intersect  at  O  at 
an  angle  of  78°  3'.     The  lines  AO,  BO,  CO,  and  DO  are  27.5,  31.8, 
58.5  and  63.2  feet  respectively.     What  is  the  area  of  the  field  ? 

35.  Two  sides  of  a  triangle  are  b  =  302  and  c  —  40.8.     Find  the 
angle  A  so  that  the  triangle  may  have  the  same  area  as  the  triangle 
whose  sides  are  62,  51  and  30.     If  b  were  30.2  and  c  were  40.8  could 
A  be  found  ?     Why  ? 

36.  Two  vessels  start  from  the  same  point  and  sail,  one  northeast 
at  the  rate  of  6  miles  per  hour,  and  the  other  east. 30°  south  at  the 
rate  of  8  miles  per  hour.     How  far  apart  will  the  ships  be  after  2|  hours  ? 

37.  A  submarine  in  submerging  drifts  back  5  feet  for  every  20  feet 
it  sinks.     After  the  submarine  has  sunk  vertically  300  yards,  at  what 
angle  must  a  torpedo  be  shot  from  a  cruiser  one  mile  away  to  hit  the 
submarine,  if  the  latter  drifts  away  from  the  cruiser  ? 

38.  A  post  6  feet  high  casts  a  shadow  10  feet  long.     What  is  the 
length  of  a  flagpole  that  casts  a  shadow  60  feet  long  if  the  pole  makes 
an  angle  of  82°  with  the  horizontal  on  the  side  away  from  the  sun  ? 

39.  In  problem  38  find  the  length  of  the  flagpole  if  the  angle  made 
with  the  horizontal  is  82°  on  the  side  towards  the  sun. 

40.  Two  yachts  begin  a  race  by  sailing  from  a  point  A,  along  the 
windward  leg  of  the   course  in  the  direction  northeast  until  they 
reach  a  buoy  B.    They  then  sail  before  the  wind,  east  32°  south, 
until  they  reach  a  point  C,  5  miles  east  along  a  straight  coast  from  A. 
The  first  yacht  sails  to  windward  5  miles  per  hour,  and  before  the 
wind  6.5  miles  per  hour  ;  the  second  5.8  miles  per  hour  to  windward 
and  6  miles  before  the  wind.     Which  yacht  wins  the  race  and  by  how 
much? 

41.  A  triangular  beach  lot  has  a  frontage  on  the  sea  of  100  yards. 
The  boundary  lines  running  from  the  beach  make,  on  the  inner  side 
of  the  lot,  angles  of  60°  and  50°  respectively  with  the  shore  line. 
How  must  a  line  be  drawn  from  the  middle  point  of  the  shore  line  to 
form  two  equal  lots  ? 


VI,  §63]      SOLUTION   OF  GENERAL  TRIANGLES        83 


42.  A  point  A,  on  the  south  bank  of  a  river  1.5  miles  broad  and 
flowing  due  east  is  to  be  connected  by  bridge  and  road'  with  a  town  U, 
3  miles  back  in  a  straight  line  from 

the  north  bank  of  the  river.  It  is 
found  that  the  bridge  can  be  built 
to  a  point,  C,  on  the  farther  bank 
lying  north  22°  west  from  J.,  or  to 
D  lying  north  41°  east  from  A. 
The  town  1}  lies  north  12°  east  from 
A.  If  the  bridge  costs  $  2000  per 
mile  to  build  and  the  road  $  500 
per  mile,  which  route  is  the  more  w 
economical  and  by  how  much  ? 
(See  Fig.) 

43.  The  distances  of  a  point  (7,  on  the  far  side  of  a  river  from  two 
points  A  and  B  on  the  near  side,  are  to  be  found  but  can  not  be  directly 

measured.     In  the  direction    CA   a   distance 
/  \  AD  =  150  feet  is  measured,  and  in  the  direction 

CB  a  distance  BE  =  250  feet.  The  distance 
from  A  to  B  is  279.5  feet,  and  by  measurement 
it  is  found  that  BD  =  315.8  feet,  DE  =  498.7 
feet.  How  far  is  C  from  A  and  B  ?  (See  Fig.) 

44.   The  distance  from  ^ 

a  point,  A,  on  the  coast  to  //  • 

a  lighthouse,  i,  is  to  be 


B 


found.     A  straight  line  is  run  from  A  along  the 

coast,  and  on  the  line  two  points,  B  and  (7,  are 

taken  from  which  the  lighthouse  is  visible.    By 

measurement  it  is  found  that  AB  =  236.7  feet, 

BC  =  215.9  feet,  the  angle  ABL  =  142°  37',  the      / 

angle  ACL  =  76°  14'.     How  far  is  the  lighthouse  / 

from  A?    (See  Fig.)  A 

45.   On  the  north  side  of  a  river  lie  two  points  A  and  B  both  of 

which  can  be  seen  from  (7,  and  from  no  other  point,  on  the  south 
side  of  the  stream.  From  a  point  D,  whose  dis- 
tance from  C  is  425.3  feet,  A  and  C  are  sighted. 
It  is  found  that  the  angle  ADC  =  37°  15',  and 
the  angle  A  CD  =  42°  35'.  From  another  point 
whose  distance  from  C  is  405.4  feet,  and 
from  which  B  and  C  are  visible,  the  angles 

CEB  =  53°  15',  and  ECB  =  58°  5'  are  measured.     The  angle  ACB  is 

also  measured  and  found  to  be  65°  11'.     What  is  the  distance  from 

AtoB?    (See  Fig.) 


CHAPTER  VII 
THE   SOLUTION  OF  TRIGONOMETRIC  EQUATIONS 

64.  The  trigonometric  equations  hitherto  dealt  with  have 
been  identical  equations ;   equations,  that  is,  true  for  any 
values  of  the  variables  involved.     We  shall  now  deal  with 
trigonometric  equations  which  are  not  identities,  and  shall 
examine  the  methods  by  which  such  equations  are  solved. 
No  methods  applicable  to  all  such  equations  can  be  given, 
but  methods  applicable  to  several  important  classes  will  be 
discussed.     In  general  it  may  be  said  that  all  such  equa- 
tions are  algebraic  in  form,  the  one  difference  being  that 
now  the  unknown  quantities  are   the   trigonometric   func- 
tions, sine,  tangent,  etc.,  or,  occasionally,  the  inverse  func- 
tions.    Therefore,  all  methods  applicable  to  the  solution  of 
algebraic  equations  are  applicable  to  the  solution  of  trigo- 
nometric equations.     Moreover,  in  the  case  of  trigonometric 
equations   we    have    the  various    fundamental    identities, 
treated  in  former  chapters,  which  being  true  for  all  values 
of  the  variables  involved  can  be  used  in  connection  with 
any  equation  whose  solution  is  desired. 

65.  For  example,  given  the  equation 

2  sin2  x  —  cos2  x  +  |  =  0 

to  find  the  value  of  x. 

In  form  this  is  an  algebraic,  quadratic  equation  in  two 
unknowns,  sin  x  and  cos  x.  To  find  the  values  of  two  un- 
knowns we  must  have  two  consistent  and  independent 
equations.  But  we  also  know  that  cos2  x  =  1  —  sin2  x. 
Therefore,  our  equation  may  be  written 

84 


VII,  §  65]        TRIGONOMETRIC   EQUATIONS  85 

2  sin2  x  -(1  —  sin2  x)  =  —  £, 

3  sin2  x  =  |, 

whence  sin  #  =  ±  -£, 

and  it*  =  sin"1  (  ±  £). 

The   principal   values  of  x  are,  therefore,   ±  -  and  the 

6 

general  values  are 


or  we  may  proceed  thus  : 

Given  2  sin2  a*  —  cos2  x  =  —  J. 

We  know  sin2  #  +  cos2  x  =  1. 

Adding  the  two  equations, 

3  sin2  x  =  },  etc. 

Example  2.     Solve  the  equation 
cos  x  —  V3  sin  x  +  1  =  0. 


For  sin  x  substitute  Vl  —  cos2  x. 

Then 

cos  x  —  V3  •  Vl  —  cos2  x  +  1  =  0, 

cos  a;  +  1  =  V3  •  Vl  —  cos2  x, 
cos2  x  +  2  cos  a;  +  1  =  3  —  3  cos2  #, 
2  cos2  a;  -j-  cos  x  —  1  =  0, 
a  quadratic  equation  in  cos  x  whose  roots  are 

cos  x  =  —  1  or  i. 
Therefore, 

a  =  cos-1  (-  1)=  2  WTT  -h  TT  =(2  n  -f-  I)TT, 

and  x  =  cos"1  ( -  ]=  2  WTT  ±  -• 

V2  3 


86  PLANE    TRIGONOMETRY  [VII,  §  65 

These  roots,  as  in  every  case,  must  be  tested  by  substitution 

in  the  original  equation.     It  is  found  that  2  n-n-  —  -  does  not 

3 

satisfy  the  equation,  while  the  other  two  values  do.     The 
roots  are,  therefore, 

x=(2n  +  V)7r  and   2mr  +-• 

3 

Another  method  of  solving  the  last  equation  is  as  follows: 
Given  cos  x  —  V3  sin  x  =  —  1. 

Divide  by  2, 


But 

Therefore  we  may  write, 

sin 


cos     cos  x  —  sin  -  sin  x  =  —  - 


Whence  x  +  -  =  cosY-  -"\  =  2  mr  ±  ?-?, 

3  \     2J  3 

and  x  =  (2  n  -  I)TT  or  2  mr  +  -• 

3 

Note  that  the  two  general  solutions  (2  n  +  I)TT  and 
(2  n  —  !)TT  are  identical  since  each  represents  any  odd 
multiple  of  TT. 

66.  Special  Types  of  Equations.  This  last  solution  is  an 
example  of  the  type  of  equation 


1.  a  cos  x  +  b  sin  x  =  c,  c<i 


VII,  §  66]        TRIGONOMETRIC   EQUATIONS  87 


To  solve  equations  of  this  type  divide  by  Va2  -f  b2. 
a        cos  x  H  --  -=-—.  sin  x  = 


Va2  +  b2  Va2  -f  62  Va2  + 

—   a        =  cos  a  and  — ==  =  sin  a 


since 


cos2  a  +  sin2  a  =  (       a       Y  +  f —  -Y  =  1. 

Wa2  4-  62/       Wa2  +  6V 

Therefore  we  may  write 


c 
cos  a  cos  x  4-  sin  a  sin  x  = 


cos  (ic  —  a)  = 


a;  —  a  =  cos"1  —   °        =  2  TITT  ±  cos"1  -  -  -  , 


Va2  +  62  Va2 

x» 

and  ic  =  2  nTr  ±  cos"1  -  —  -f-  a. 

Va2  +  62 

Another  type  of  equation  is 
2.  tan  a0  =  cot  66     or     sin  a9  =  cos  &9. 

We  may  put 
cot  bB  =  tan  (?-  —  bO\  ;  sin  aO  =  cos  (?•  —  a6  \ 

Therefore 
tan  aO  =  tan  f-  —  b6\  cos  bO  =  cos  (-  —  aO\ 

bO  =  2  mr  ±  (j  -  a$\ 


88  PLANE   TRIGONOMETRY  [VII,  §66 

or  •*  2tt7r±? 


Example.     Given  tan  3  0  =  cot  2  0,  find  6. 


Therefore,  tan  3  0  =  tan  (?  -  2  A 

\*          / 


5  10 

A  third  type  of  equation  is 

3.  sin  ax  +  sin  bx  +  sin  ex  =  0,  . 

cos  ax  4-  cos  bx  +  cos  ex  —  0, 
cos  ax  4-  cos  bx  4-  sin  CAT  =  0, 
sin  ax  +  sin  bx  4-  cos  a:  =  0. 

To  solve  equations  of  this  type,  formulae  (19)  and  (20) 
are  used. 

Example.     Solve  the  equation  sin  5  x—  sin  3  x  4- sin  #=0. 
We  may  write 

sin  5  x  —  sin  3  x  =  2  cos  i  (5  x  +  3  x)  sin  £  (5  x  —  3  a?) 
=  2  cos  4  a;  sin  x. 

Therefore       2  cos  4  x  sin  x  +  sin  a:  =  0, 

sin  x  (2  cos  4  a?  4- 1)=  0. 
Whence, 

sin  x  =  0  or     2  cos  4  a;  4- 1  =  0 

x  =  sin"1 0  4  a;  =  cos"1  (—  ^-) 

x  =  sin"1  0  =  nir  4  «  =  cos"1  (—  |)=  2  nir  ±  — ^ 

97/7T    .     7T 


VII,  §  67]        TRIGONOMETRIC   EQUATIONS  89 

67.    Simultaneous  equations  involving  trigonometric  func- 
tions can  in  many  cases  be  solved. 

Example  1.      Given  y  =  1  —  cos  x,  y  =  1  +  sin  x,   find  x 
and  y. 

We  have  1  +  sin  x  =  1  —  cos  xy 

sin  x  =  —  cos  x, 
tan  x  =  —  1. 

o 

a  =  tan-i  (—  1)  =  mr  -f  —^, 
4 

and  2/=l  +  sina  =  l-cosa=l±-tr 

V2* 

Example  2.     Given  r  cos  (  0  —  -j=a,  rcos/0—  -)  =  a, 
find  r  and  6.  \ 

We  have 

r  cos  f  (f  —  ~  }  = 


cos    ^ }  = 


r  =  asec  f  0  —  -   = 


a  sec    riTT--^    = 


When  n  is  even,  sec  [  nir  — ^  )  =  —  sec  ^ 
\          12/  12 


sec 


90  PLANE    TRIGONOMETRY  [VII,  §67 

When  n  is  odd: 

sec  f mr  ~  jQ  -  sec  f *»  +  JTJ)  =  ~  sec  J7T 

Therefore,  r  =  —  a  sec  — . 

12 

68.    Equations   Involving  Inverse  Trigonometric  Functions 

may,  in  general,  be  solved  by  transforming  to  other,  equiv- 
alent equations  involving  the  direct  functions.  The  method 
of  solution  is  illustrated  by  the  following  examples. 

Example  1.     Solve  the  equation  2  tan"1  x  =  cot"1  a?. 
We  have         cot  (2  tan^1  x)  =  cot  (cot"1  x) 


2  cot  (tan"1  x) 

That  is,  —t —  =x  or  3 x2  =  1 ;  whence  x=±  —• 
2  v3 

x 
Example  2.     Solve  the  equation  cos"1  x  +  sin"1 2  a;  =  0. 

We  have  .    .  .     .  0   x      A 

sin  (cos"1  a;  +  sin"1 2  a;)  =  0, 


or 


sin  (cos"1  #)  •  cos  (sin"1 2  x)  -\-  cos  (cos  J  x)  •  sin  (sin"1 2  x)  =  0. 


That  is,  ±  Vl  —  a2  •  Vl  -  4  a2  +  x  •  2  a?  =  0. 

Whence,(l-z2)(l-4o;2)  =  4a4,  5o;2=  1,  x=±  — 

V5 

A  second  method  of  solving  example  2  is  as  follows : 

cos"1  x  =  —  sin"1 2  #, 
sin  (cos"1  x)  =  sin  (  —  sin"1 
VI  -  a2  =  -  2  a?, 
5  a;2  =  1, 


VII,  §68]        TRIGONOMETRIC   EQUATIONS 


91 


V5 

In  every  case  the  values  must  be  checked  by  substitution  in 
the  original  equation. 


FIG.  30. 

It  is  often  convenient,  in  dealing  with  inverse  functions, 
,to  assume  the  angle  whose  function  is  given  and  to  construct 
a  figure  to  show  the  values  of  the  remaining  functions. 
Thus,  in  example  1,  we  wish  to  find  cot  (tan-1  x).  Let 
tan"1  x  =  a  and  construct  the  angle  a,  Fig.  30,  with  ordinate 
equal  to  ±  x  and  abscissa  equal  to  ±  1.  The  distance  is 


FIG.  31. 


then    Vl  4- a2,  and  all   the   functions   are   readily  found. 

Thus,  cot  (tan"1  #)  =-•     Similarly,  in  example  2,  we  wish 
x 

to  find  sin  (cos"1  x)  and  cos  (sin"1 2  x).  By  figure  31  we  see 
assuming  cos"1  x  =  a  and  sin"1 2  x  =  b,  that  sin  (cos"1  x)  = 
±  Vl  —  x2  and  cos  (sin"1 2  x)  =  ±  Vl  —  4  a;2. 


92  PLANE    TRIGONOMETRY  [VII,  §  68 

EXAMPLES 

Solve  the  following  equations  : 

1.  sin5x  — sin3x+ sinx  =  0. 

2.  cos  0  +  cos  2  6  +  cos  3  0  =  0. 

3.  sin  4  x  —  sin  2  x  —  cos  3  x  =  0. 

4.  6  sin  0  +  cos  0  =  2.  9.   tan  x  =  cos  x. 

5.  2  sin  0  +  cos  6  =  2.  '  10.   cos  2  x  =  £  cos  x. 

6.  sin2x  =  sinx.  11.   sin 2  0cos20  +  2sin0  =  0. 

7.  cos2x  =  sinx.  12.   sin4x— 2  sin  x  cos  2  x  =  0. 

8.  sin  3  6  =  cos  0.  13.    sin  4  6  =  cos  2  0. 

14.  cos20  =  sin20  —  1. 

15.  cos  (x  —  a)  cos  x  —  sin  (x  —  a)  sin  x  =  0. 

16.  sec2  x  =  3  esc2  x. 

17.  sin  0  cos  0  -  sin  f-  -  0^  cos  I-  -  0^  =  0. 

18.  27  esc  0  cot  0  =  8  sec  0  tan  0. 

19.  25  sin  0-12.8  esc2  0  =  0. 

20.  cos3  0  —  2  sin2  0  cos  0  =  0. 

Solve  the  following  simultaneous  equations  for  x  and  ?/,  or  r  and  0. 

21.  i/  =  l  — cos2x,  27.    r  =  3sin0 -f  2cos0, 
y  =  1  -f  sin  2  x.  r  =  3  cos  0  +  2  sin  0. 

28'     2 


99      „. 

~~2'  2/2sin2x=l. 

o0  on 

r  =  esc2  -  •  29.   y  =  sm  x, 

?/  =  sin  (x  -f-  a). 

23.   r=  a  sin  0,  /          .       o  n 

o/\    /  /»     ir\      o  a 

r  =  a  sin  2  0.  <du- 


24.  y  = 

2/  =  2asinxtanx. 

25.  y  cosx  =  2  a,  r2  =  sin  3  0. 


32. 

26.    r2  sin  2  0  =  1,  r  =  2  sec  0. 

33.    r*  sin  2  0  =  4, 

r2  =  16  sin  2  0. 


VII,  §68]        TRIGONOMETRIC   EQUATIONS  93 

Solve  the  following  equations  : 

34.  sin-1  x  =  sin-1  a  +  sin-1  b.  37.  sin-1  x  =  cos-1  (—  x). 

35.  sin-ix^  sin-12x  =  -«  38.  tan~12x+tan~13x=  —  • 

3  4 

.  _j4       .   _18_?r  39  t     -1     —  2       -ix 

L    -              -  _  -  •  .  an    x  _     cos    -  • 


40.   tan-1  (x  +  1)  -  cot'1  (  -^—}  =  tan'1 2 
\x-  I/ 


41. 


-  1       3 


42.  tan-1?—-!  +  tan-1  — *—  =  tan-i(  -  2). 

x  +  1  x-1 

Solve  the  following  equations,  finding  only  the  principal  values  of 
angles : 

43.  cos  5  x  —  cos  3  x  +  sin  x  =  0.        49.   y  =  tan  2  x, 

45.  cos  3  x  =  cos  x.  50.    r  =  sin  0, 

46.  cotx  =  cosx.  r  =  s: 

47.  y=  1  +cos2x, 

i      „•„  o  «.  61.    sin  4  0  =  cos  6. 

y  —  L  —  sin  4  x. 

48.  r  =  acos0,  62'   3sin»  +  cos«  =  2. 
r  =  a  sin  2  0. 


SPHERICAL   TRIGONOMETRY 


CHAPTER   VIII 
FUNDAMENTAL   RELATIONS 

69.  Spherical    trigonometry    deals    with,    the    relations 
among  the  sides  and  angles  of  a  spherical   triangle ;  that 
is,  of  a  portion  of  the  surface  of  a  sphere  bounded  by  the 
intersecting  arcs  of  three  great  circles.     It  deals  also  with 
the  computation  of  unknown  parts  of  such  a  triangle  from 
parts  which  are  known,  the  process  being  called,  as  in  plane 
trigonometry,  the  solution  of  the  triangle.     The  sides  of  a 
spherical  triangle,  being  arcs  of  circles,  are  expressed  in 
degrees,  minutes,  and  seconds,  and,  as  is   customary,  we 
shall  consider  only  those  triangles  in  which  each  part  (angle 
or  side)  is  less  than  one  hundred  and  eighty  degrees. 

70.  Law  of  Cosines.     There  is  one  theorem,  the  law  of 
cosines,  which  may  be  called  the  fundamental  theorem  of 


FIG.  32. 

spherical  trigonometry,  because  by  means  of  the  theorem 
any  spherical  triangle  may  be  solved  when  three  of  its  parts 
are  known.  We  shall  proceed  to  prove  the  Law  of  Cosines. 
Let  ABC,  Fig.  32,  be  a  spherical  triangle  on  a  sphere 
whose  center  is  0,  and  let  the  sides  b  and  c  be  less  than  90°. 

94 


VIII,  §70]        FUNDAMENTAL   RELATIONS  95 

Through  any  point,  4',  on  OA  pass  a  plane  perpendicular 
to  OA  cutting  the  planes  04(7,  GAB,  and  OBC,  in  A'C', 
A'B',  and  B'C',  respectively.  Then  the  angle  B'A'O'  is  the 
measure  of  the  diedral  angle  B-OA-C  and,  therefore,  of 
the  spherical  angle  A.  Also,  by  the  construction,  the  angles 
OA'B'  and  OA'C'  are  right  angles.  In  the  triangle  A'B'C' 


C7^2  -  2  B'A'  •  C'A'  cos  A, 

and  in  the  triangle  B'OC' 

B'C12  =  Br02+CrOz-2B'0'  C"0cosa. 
Whence 

Wcf+Cfff-ZB'O  >  C'Ocosa 

=  BTA'2  +  CM72  -  2  £'4'  -  CM'  cos  A, 
or 

2£'O-  C'Ocosa 

=  WO2  -  WA'2  +  (70*  -  C'A*2  +  2  B'A'  •  C'A'  cos  A 

But  B'OA'  and  C'OA'  are  right  triangles,  and  therefore, 


We  then  have 

B'O  •  C'Ocosa  =  04'2+  B'A'  •  C"^'  cos  A, 
or  04'      04'  ,B'4'      0'4' 


But 

f)  A' 


cos  J304  =  cos  c,    -^p  =  cos  40C  =  cos  6, 


, 
BO  CO 


=  sin  504  =  sin  c,          -  =  sin  40C  =  sin  b. 
BO  CO 

Hence  cos  a  =  cos  6  cos  c  4-  sin  6  sin  c  cos  4, 

which  is  the  law  of  cosines. 


96  SPHERICAL   TRIGONOMETRY        [VIII,  §70 

In  the  above  demonstration  the  sides  b  and  c  were  taken 
less  than  90°  in  order  that  the  construction  of  the  right  tri- 
angles B'OA'  and  C'OA'  might  be  possible.  The  resulting 
theorem,  however,  is  true  in  all  cases.  Let  us  assume 
90°  <  b  <  180°  and  90°  <  c  <  180°.  Then,  Fig.  33,  pro- 
duce the  arcs  AB  and  AC  to  meet  in  A',  thus  forming  a 
lune.  In  the  triangle  A'BC,  b'  and  c'  are  less  than  90°. 


FIG.  34. 

The  law  of  cosines  is,  therefore,  true  for  the  triangle  A'BC, 
so  that,  since  A'  =  A, 

cos  a  =  cos  b'  cos  c'  -f-  sin  b'  sin  c'  cos  A. 
But  b1  =  180° -b  and  c'  =  180°-c. 

Whence 

cos  a  =  cos  (180°  —  b)  cos  (180°  —  c) 

+  sin  (180°  -  b)  sin  (180°  —  c)  cos  A, 

or,  cos  a  =  cos  b  cos  c  -f-  sin  b  sin  c  cos  A.  Q.E.D. 

Again,  let  b  <  90°  and  90°  <  c  <  180°.  Produce  the  arcs 
BA  and  BC,  Fig.  34,  to  meet  in  B',  thus  forming  a  lune. 
Then,  in  the  triangle  AB'C,  b  <  90°  and  cf  <  90°,  and, 

therefore,         cos  a' =  cos  6  cos  c' +  sin  6  sin  c' cos  CAB'. 
But 

a'  =  180°  -  a,  c'=  180°  -  c,  and  CAB'  =  180°  -  A. 

Hence 

cos  (180°  —  a) 

=cos  b  cos  (180°  -  c)  +  sin  b  sin  (180°  -  c)  cos  (180°  -  A), 


VIII,  §71]        FUNDAMENTAL   RELATIONS  97 

or,  cos  a  =  cos  b  cos  c  -f  sin  b  sin  c  cos  -4, 

which  proves  the  law  of  cosines  for  all  cases. 

We  thus  have  the  three  fundamental  equations,  the  law 

of  cosines : 

cos  a  =  cos  b  cos  c  +  sin  b  sin  c  cos  A, 

cos  6  =  cos  c  cos  a  -f-  sin  c  sin  a  cos  JB,  (37) 

cos  c  =  cos  a  cos  b  -f-  sin  a  sin  &  cos  C, 

by  means  of  which  any  spherical  triangle  may  be  solved. 
For  example,  given  a  =  60°,  b  =  70°,  A  =  65°. 

We  have 

cos  60°  =  cos  70°  cos  c  +  sin  70°  sin  c  cos  65,° 
°r>  .500  =  .342  cos  c  +  .940  sin  c  x  .423, 

.342  cos  c  +  .398  sin  c  =  .500, 

.342  cos  c  +  .398  sin  c .500 

V(.342)2+(.398)2      "  V(.342)2  + (.398)2 ' 

.342  ,  .398    .  .500 

__cosc+—  sine-— , 

.651  cos  c  +  .758  sin  c  =  .952, 

.651  =  cos  49.4°,  .758  =  sin  49.4°. 
Therefore, 

cos  (c- 49.4°)  =  .952, 

c  -  49.4°  =  cos-1 .952  =  18.2°  and  c  =  67.6°. 

Similarly  the  other  parts  may  be  found.  The  equations 
are  not,  however,  adapted  to  logarithmic  computation,  so 
that  for  practical  use,  as  will  presently  be  shown,  they 
must  be  transformed  in  various  ways. 

71.   Law   of  Cosines   Applied  to  the  Polar  Triangle.     The 

law  of  cosines,  being  true  for  any  triangle,  is  true  for  the 
polar  triangle  of  ABC.     Therefore,  denoting  the  six  parts 
n 


98  SPHERICAL  TRIGONOMETRY        [VIII,  §71 

of  the  polar  triangle  by  the  same  letters  accented,  we  have 

cos  a'  =  cos  b'  cos  c'  -f-  sin  &'  sin  c'  cos  A'. 
But 

a'  =  180°  -A,b'  =  180°  -  B,  c'  =  180°  -  C,  A'  =  180°  -  a. 

Whence, 

cos  (180°  -A)=  cos  (180°-  B)  cos  (180°-  (7)  -f  sin  (180°-  B) 

sin  (180°  -  <7)  •  cos  (180°  -  a), 
or 

—  cos  A  =  cos  B  cos  (7  —  sin  B  sin  (7  cos  a, 

so  that  the  truth  of  the  three  following  equations  is  obvious  : 

cos  A  =  —  cos  B  cos  C  •+-  sin  B  sin  C  cos  a, 

cos  B  =  —  cos  (7  cos  .4  -h  sin  (7  sin  A  cos  6,          (38) 

cos  (7  =  —  cos  ^4  cos  B  +  sin  ^4  sin  B  cos  c. 

72.  Law  of  Sines.  Another  theorem  of  importance  in  the 
solution  of  spherical  triangles,  known  as  the  law  of  sines,  is 
as  follows :  In  any  spherical  triangle  the  sines  of  the  sides  are 
proportional  to  the  sines  of  the  opposite  angles.  That  is, 

sin  a :  sinb:  sin  c  —  sin  A :  sin  B:  sin  C.  (39) 

From  equations  (37)  we  have, 
A      cos  a  —  cos  b  cos  c 

COS  A  = : : —        — . 

sin  b  sin  c 
Whence 

1  -  cos2  A  =  1  -  (CQS  «  -  cos  6  cos  c)2 

sin2  b  sin2  c 
or, 

.  2   .  _  sin2  b  sin2  c  —  (cos  a  —  cos  b  cos  c)2 
sin2  b  sin2  c 

_  (1  —  cos2  6)(1  —  cos2  c)  —  (cos  a  —  cos  6  cos  c)2 
sin2  b  sin2  c 

_  1  —  cos2  6  —  cos2  c  —  cos2  a +  2  cos  a  cos  6  cos  c 
sin2  &  sin2  c 


VIII,  §72]        FUNDAMENTAL   RELATIONS  99 

Whence, 

sin2  A  _  1  —  cos2  a  —  cos2  6  —  cos2  c  +  2  cos  a  cos  6  cos  c 
sin2  a                                 sin2  a  sin2  b  sin2  c 
or,  

sin  A  _  VI  —  cos2  a  —  cos2  6  —  cos2  c+2  cos  a  cos  6  cos  c 

sin  a  sin  a  sin  6  sin  c 

where  the  positive  sign  is  taken  because  A  and  a  are  each 
less  than  180°.  The  right-hand  member  of  this  expression  is 
symmetric  in  a,  6,  and  c,  so  that  if  we  started  with  cos  B  or 
cos  C  instead  of  with  cos  A,  the  final  result  for  the  right- 
hand  member  would  be  identical  with  that  written  above. 
Therefore,  obviously,  we  have 

sin  A  _  sin  B  __  sin  C 
sin  a       sin  b       sin  c ' 

the  law  of  sines  which  was  to  be  proved. 


CHAPTER   IX 
THE   SOLUTION  OF  RIGHT  SPHERICAL  TRIANGLES 

73.  Special  Formulae  for  Right  Triangles.  If  we  let  C  be 
the  right  angle  in  a  right  spherical  triangle,  and  put  C  =  90° 
in  the  third  equation  of  (37),  we  have 

cos  c  =  cos  a  cos  b.  (40) 

The  third  equation  of  (38)  gives 

0  =  —  cos  A  cos  B  +  sin  A  sin  B  cos  c, 
or,  cos  c  =  cot  A  cot  B.  (41) 

The  first  two  equations  of  (38)  give 

cos  A  =  sin  B  cos  a, 
cos  B  —  sin  A  cos  b. 

Using  the  proportions  of  (39)  when  C  =  90°,  we  have 

sin  A  _  sin  j?  __     1 
sin  a      sin  6      sin  c 
Whence, 

sin  a  =  sin  A  sin  c, 

(43) 
sin  b  =  sin  B  sin  c. 


From  (42)  by  (43)  we  have 
or,  by  (40) 


cos  A  =  sin  B  cos  a  =      — .  cos  a, 
sine 


A      sin  6    cos  c      tan 
cos  A  = 


sm  c    cos  b     tan  c 
100 


IX,  §74]       RIGHT   SPHERICAL   TRIANGLES  101 

Similarly, 

(44) 


tanc 
By  (43),  (44),  and  (40), 


sng 


sin  A      sin  c      sin  a    tan  c 

tan  A  =  -    -;  =  z  —  r  =  -:    -  •  7  —  r 

cos  A     tan  6     sin  c    tan  b 

tanc 
sin  a  sin  a  tan  a 


tan  b  cos  c     tan  &  •  cos  a  cos  6      sin  b 

In  the  same  way, 

(46) 


sin  a 


74.   The  formulae  (40)  to  (45)  may  be  assembled,  in  a 
slightly  different  form,  as  follows  : 


sin  a  n     sin  b 

sin  A  =  —  —  •  sin  5  = 

sin  c  sin  c 


tan  c  tan  c 


(46) 


sin  b  sin  a 

cos  A  =  cos  a  sin  B.  cos  B  =  cos  b  sin  4. 

cos  c  =  cos  a  cos  &.  cos  c  =  cot  ^4  cot  B. 

A  device,  known  as  Napier's  Rules,  was  formulated  by 
Napier  to  facilitate  the  remembering  of  the  above  formulae. 
Let  us  take  for  the  Jive  parts  of  a  right  triangle  the  sides 
a  and  6,  and  the  complements  of  A,  B,  and  c.  These  five 
parts,  Fig.  35,  arrange  themselves  so  that  each  is  a  middle 
to  two  adjacent  parts  and  a  middle  to  two  opposite  parts. 

NAPIER'S  RULES  state 

I.  The  sine  of  the  middle  part  equals  the  product  of  the 
tangents  of  the  adjacent  parts. 


102  SPHERICAL   TRIGONOMETRY          [IX,  §74 

II.    The  sine  of  the  middle  part  equals  the  product  of  the 
cosines  of  the  opposite  parts. 


co-B 


co-A 


By  applying  these  rules  to  the  various  parts  all  the 
formulae  of  (46)  may  be  obtained.  Thus,  for  example, 

sin  (co-^1)  =  tan  b  •  tan  (co-c). 
That  is, 

sin  (90°  -  A)  =  tan  6  •  tan  (90°  -  c), 

or,  cos  A  =  tan  b  cot  c  =  —5-  • 

tanc 

75.  Rules  for  Solution.  In  a  right  triangle,  the  right 
angle  being  always  known,  only  two  other  parts  need  be 
known  to  solve  the  triangle.  To  solve  a  right  triangle  by 
means  of  the  formulae  (46)  we  have,  therefore,  the  general 
rule :  Select  that  equation  which  involves  the  two  known  parts 
and  one  unknown  part. 

The  algebraic  signs  of  the  functions  must  be  carefully 
noted  in  order  to  determine  the  sign  of  the  resulting  func- 
tion and  thereby  the  angle.  If  the  part  to  be  found  is  got 
from  a  cosine,  tangent,  or  cotangent  there  is  no  ambiguity, 
for  if  these  functions  are  plus  the  part  will  have  a  value 
less  than  90°.  If  they  are  minus  the  part  will  have  for  its 
value  the  supplement  of  the  angle  found  from  the  tables  of 
trigonometric  functions. 

On  the  other  hand,  if  the  unknown  part  is  determined  by 
a  sine,  the  sine  being  positive  for  all  angles  between  0°  and 
180°,  the  value  may  be  either  that  got  from  the  tables  or  its 


IX,  §76]       RIGHT   SPHERICAL   TRIANGLES  103 

supplement.     In  general  both  solutions  must  be  used  unless 
the  ambiguity  can  be  removed  by  the  following  laws  : 

1.  If  the  sides  adjacent  to  the  right  angle  are  in  the  same 
quadrant,  the  hypotenuse  is  less  than  90°  ;  if  they  are  in  dif- 
ferent quadrants,  the  hypotenuse  is  greater  than  90°. 

2.  An  angle  and  its  opposite  side  are  in  the  same  quadrant. 

PROOF  OF  LAW  1.     By  (40)  cos  c  =  cos  a  cos  b. 

Let  a<90°  and  b  <  90°.  Then  cosa  =  ±,  cos&  =  ±, 
and  cos  c  =  (  ±  )(  ±  )  =  -f  .  Therefore,  c  <  90°.  Again,  let 
a  ^90°,  and  b  ^90°.  Then  cos  a  =  ±,  cos&  =  =F,  cose 
-,  and  c>90°. 


PROOF  OF  LAW  2.     By  (45)  sin  b  = 


tan^L 


Since  sin  b  is  necessarily  positive,  it  follows  that  tan  a 
and  tan  A  are  both  plus  or  both  minus.  Therefore  a  and  A 
are  each  less  than  90°  or  each  greater  than  90°. 

76.  The  solution  of  right  triangles  is  illustrated  by  the 
following  examples  : 

Example  1.     Given  A  =  33°  50',  b  =  108°,  find  B,  a,  and  c. 
From  the  formulae  (46)  we  select 

tan  a        A   tan  b 

tan  A  =  -r—  -  ,    cos  A  =  —  —  ,     cos  B  =  cos  6  sin  A. 
sin  b  tan  c 

or, 

tan  a  =  tan  A  sin  6,    cot  c  =  cos  A  cot  b,  cos  B  =  sin  A  cos  b. 

+log  tan  A  =  9.8263  +log  cos  A  =  9.9194 

+log  sin  b    =  9.9782  "log  cot  b  =9.5118 

+log  tan  a  =  9.8045  "log  cot  c'  =  9.4312 

c'  =  74°  54' 
a  =  32°  31'  c  =105°  6' 


104  SPHERICAL   TRIGONOMETRY          [IX,  §76 

4log  sin  A  =  9.7457 
-log  cos  b  =  9.4900 
-log  cos  B'  =  9.2357 

B'  =  80°  6' 

B  =  99°  54' 

To  check  the  results  we  select  a  formula  involving  the 

three  parts  to  be  found  ;  a,  c,  and  B.     Thus  cos  B  =  J£J?, 

tanc 

log  cos  B  =  log  tan  a  -f-  log  cot  c 

9.2357  =9.8045  +  9.4312  CHECK. 

Example  2.     Given  a  =  47°  30',  c  =  120°  20',  find  A,  B, 
and  6. 


We  have 


sin  a  T.     tan  a  ',       cos  c 

—  —  ,        cos5  =  —   —  ,        cos  b  =  —  —  . 
sin  c  tan  c  cos  a 


log  sin  a  =9.8676  +log  tan  a  =0.0379  ~log  cos  c  =9.7033 
log  sine  =9.9361  "log  tanc  =0.2327  +log  cos  a  =9.8297 
log  sin  A  =9.9315  -log  cos  £'=9.8052  -log  cos  &'=9.8736 

£'=50°  19'  &'=41°  38' 

A  =  58°40'  B  =  129°  41'  b  =  138°  22 

CHECK.  cos  B  =  cos  6  sin  A 

log  cos  B  =  log  cos  b  +  log  sin  A 
9.8052  =  9.8736  +  9.9315  =  9.8051. 

NOTE.     The  value  of  A  less  than  90°  is  taken  by  virtue 
of  law  2,  Art.  75. 

Example  3.    Given  B  =  105°  59',  b  =  128°  33',  find  A,  o, 
and  c. 

We  have, 
cos  B  =  cos  b  sin  A,  tan 


sn  a  sin  c 


IX,  §76]       RIGHT    SPHERICAL   TRIANGLES  105 


or 


sin  a  =  tan  b  cot  B, 


,  ,  --. 

cos  b  sm  jB 

-log  cos  B  =9.4399     ~log  cot  B=  9.4570     log  sin  b  =9.8932 

-log  cos  b  =9.7946     -log  tan  b  =0.0986     log  sin  5=9.9828 

log  sin  A  =  9.6453      log  sin  a  =  9.5556     log  sin  c  =  9.9104 

^1=26°  14'  ai=21°  4'  Cl  =54°  27' 

A2  =  153°  46'  a2  =  158°  56'  c2  =  125°  33' 

CHECK.  sin  a  =  sin  A  sin  c. 

log  sin  a  =  log  sin  A  -f  log  sin  c 
9.5556  =  9.6453  +  9.9104  =  9.5557. 

By  law  2  both  sets  of  values  must  be  used  ;  but  by  law  1 
the  acute  value  cx  belongs  with  the  obtuse  values  of  A  and  a, 
the  obtuse  value  c2  with  the  acute  values  of  A  and  a.  Thus 
the  two  solutions  are  : 

1.  A  =  26°  14',  a  =  21°  4',  c  =  125°  33'. 

2.  .4  =  153°  46',         a  =  158°  56',         c  =  54°  27'. 

NOTE.  A  quadrantal  spherical  triangle  is  one  which  has 
a  side  equal  to  a  quadrant.  The  polar  triangle  of  a  quad- 
rantal triangle  is  right.  Therefore,  to  solve  a  quadrantal 
triangle  solve  its  polar  triangle  and  take  the  supplements 
of  the  parts  thus  found. 

EXAMPLES 

Solve  the  following  triangles  in  which  C  =  90°. 

1.  A  =  40°  13',  5.    a  =165°  19',  9.     a  =  144°  1', 

a  =  26°  25'.  c  =  46°  50'.  6  =  123°  6'. 

2.  B  =  83°15',  6.     6  =40°  49',  10.    4  =  69°  17', 

6  =  76°  46'  .  c  =  135°  40'.  B  =  51°  46'. 

3.  B  =  110°  50',  7.    a  =  21°  18',  11.  ^  =  137°  18', 

6  =  118°  30'.  6  =  49°  55'.  B  =  119°  30'. 

4.  6  =  127°  36',  8.    a  =  78°  32',  12.   A  =  71°  46', 
c  =  94°  52'.                      b  =  132°  25'.  B  =  148°  3'. 


106  SPHERICAL   TRIGONOMETRY          [IX,  §76 


13. 

A 

=  20°  34', 

19. 

A  = 

98°  17', 

25. 

A 

— 

97°  24', 

c 

=  23°  18'. 

a  = 

143°  8'. 

a 

= 

103D  12'. 

14. 

B 

=  97°  36', 

20. 

a  = 

172°  28', 

26. 

b 

= 

164°  10', 

c 

=  96°  31'. 

c  — 

124°  39'. 

c 

= 

133°  60'. 

15. 

A 

=  100°  38', 

21. 

a  = 

4°  54', 

27. 

b 

= 

34°  3', 

c 

=  51°  44'. 

b  = 

169°  27'. 

a 

= 

54°  26'. 

16. 

B 

=  59°  54', 

22. 

A  = 

76°  17', 

28. 

A 

— 

156°  30', 

a 

=  6°  60'. 

B  = 

144°  1'. 

B 

= 

104°  50'. 

17. 

B 

=  47°  34', 

23. 

B  = 

82°  43', 

29. 

A 

= 

165°  1', 

a 

=  144°  24'. 

c  = 

99°  26'. 

c 

= 

50°  30'. 

18. 

A 

=  102°  49', 

24. 

TJ  

99°  47', 

30. 

B 

= 

37°  56', 

b 

=  10°  19'. 

a  — 

26°  43'. 

a 

=r 

157°  12'. 

Solve  the  following  quadrantal  spherical  triangles  (c  =  90°)  : 

31.  ^1  =  30°  12',  33.     6  =  51°  33',  35.    A  =  159°  20', 

a  =  72°  29'.  C  =  25°  48'.  a  =  136°  30'. 

32.  £  =  118°  16',  34.   ^  =  141°  13',  36.     b  =  18°  41', 
a  =137°  57'.                  C  =  49°  35'.  -     A  =  39°  24'. 


CHAPTER  X 
THE   SOLUTION   OF    OBLIQUE    SPHERICAL   TRIANGLES 

77.  Six   Cases   may   be   enumerated   in   the   solution   of 
oblique  spherical  triangles. 

1.  Given  the  three  sides,  a,  b,  c. 

2.  Given  the  three  angles,  A,  B,  C. 

3.  Given  two  sides  and  the  included  angle,  a,  b,  C. 

4.  Given  two  angles  and  the  included  side,  A,  B,  c. 

5.  Given  two  sides  and  the  angle  opposite  one  of  them,  a, 
b,A. 

6.  Given  two  angles  and  the  side  opposite  one  of  them,  A, 
B,  a. 

We  shall  proceed  to  consider  these  cases  in  the  order 
named. 

78.  Case  1.     Given  the  three  sides.     The  law  of  cosines  is 
sufficient   to   solve   this   case,   but   the   equations   are   not 
adapted   to    logarithmic   computation.      We   therefore   de- 
velop them  as  follows : 

We  have  proved  the  formula 


—  cos^l 

2L  =\/ • 

^  1  -h  cos  A 
By  the  law  of  cosines 

^_  cos  a  —  cos  6  cos  c 

sin  b  sin  c 
107 


108  SPHERICAL   TRIGONOMETRY  [X,  §78 

Whence 

1  —  cos  A  _  sin  b  sin  c  +  cos  b  cos  c  —  cos  a 
1  -f-  cos  A     sin  b  sin  c  —  cos  b  cos  c  -f-  cos  a 

_  cos  (6  —  c)  —  cos  a  _  cos  a  —  cos  (b  —  c) 

~  cos  a  —  cos  (b  -f  c)      cos  (6  +  c)  —  cos  a 
But 

cos  a  —  cos  (b  —  c)  =  —  2  sin  |  (a  +  b  —  c)  sin  £  (a  —  b  -f-  c), 
cos(6  +  c)—  cosa  =  —  2  sin  1  (a  +  6  +  c)  sin^(6  +  c  —  a). 

Hence 

—  c)sin|(a  — 


1  +  cos  .4      sin  ^(a  +  6  +  c)  sin  ^(b-\-  c  —  a) 
Let  a  +  6  +  c  =  2s;  then  a  +  6  —  c  =  2  (s  —  c), 

a  —  6  +  c  =  2(s  —  6),  and  6  +  c-  a  =  2(s  —  a). 


Therefore,       1  —  cos  A  _  sin  (s  —  b)  sin  (s  —  c) 
1  +  cos  A          sin  s  sin  (p  —  a) 


tan^=.--.  (47) 

Similarly, 


sin 


(•-*) 

We  may  write 


sin  (s  —  a)  sin  (s  —  b)  sin  (s  —  c) 


sin  (s  —  a)       *  sin  s 

or,  putting       

/sm  (s  —  a)  sin  (s  —  b)  sin  (s  —  c) 
sins 


(48) 

sin  (s  —  c) 


X,  §78]       OBLIQUE    SPHERICAL   TRIANGLES  109 

Either  set  of  formulae  (47)  or  (48)  may  be  used  in  the  solu- 
tion of  Case  1.  If  a  check  is  desired  in  the  solution  the  law 
of  sines  may  be  so  used.  Thus,  since 

sin  A  _  sin  B  _  sin  C 
sin  a       sin  b       sin  c  ' 
it  follows  that 

log  sin  A  —  log  sin  a  =  log  sin  B  —  log  sin  b 
=  log  sin  C  —  log  sin  c. 

It  must  be  remembered,  however,  that  results  may  check 
and  still  be  incorrect.  If  they  do  not  check  they  are  wrong  ; 
if  they  check  they  may  be  right,  or  may  be  wrong,  since 
errors  may  compensate  each  other.  It  is  important  to 
check  one's  work,  but  far  more  important  to  learn,  by  care- 
ful attention,  to  work  accurately. 

Example.     Given   a  =  103°,  6  =  53°,  c  =  61°,  find  A,  B, 
and  C. 

Using  the  formulae  (48)  we  find 

8  =  i  (103°  +  53°  +  61°)  =  108°  30', 
s-a  =  5°30',  s-6  =  55°30r,  s-c  =  47°30'. 

log  sin  (s  —  a)=    8.9816 

log  sin  (s  -  b)  =    9.9160 

log  sin  (s  —  c)  =    9.8676 

log  esc  s  =    0.0230 

2)18.7882 

log&=    9.3941 

log  k  =  9.3941  log  k  =  9.3941 

log  sin  (s—a)  =  8.9816  log  sin  (s  —  b)  =  9.9160 

log  tan  £  A  =  0.4125  log  tan  |  B  =  9.4781 

%  A  =  68°  51',  I B  =  16°  44', 

A  =  137°  42',  B  =  33°  28', 


110  SPHERICAL   TRIGONOMETRY  [X,  §78 

log  k  =  9.3941 
log  sin  (s  —  c)  =  9.8676 
log  tan  ±C  =  9.5265 
iC=18°35', 
C  =  37°  10'. 
CHECK. 

log  sin  A  =  9.8280     log  sin  B  =  9.7415     log  sin  C  =  9.7811 

log  sin  a  =  9.9887     log  sin  b  =  9.9023     log  sin  c  =  9.9418 

9.8393  9.8392  9.8393 

79.  Case  2.  Given  the  three  angles.  This  case  may  be 
solved  by  the  same  formulae  that  are  used  in  Case  1,  by 
making  use  of  the  principle  of  polar  triangles.  Thus,  using 
accented  letters  to  represent  the  corresponding  parts  of  the 
polar  triangle,  we  have  a'  =  180°  -  A,  b'  =  180°  -  B,  c'  = 
180°  —  C.  Knowing  the  sides  a',  &',  c',  we  can  find  the 
angles  A',  B',  C',  as  in  Art.  78.  Then  the  sides  of  the 
original  triangle  will  be 

a  =  180°  -A',b  =  180°  -  B',  c  =  180°  -  C'. 
Example.     Given  A  =  123°,  B  =  43°,  C  =  64°,  find  a,  6,  c. 
Here          a'  =  180°  -  A  =  57°,  b'  =  137°,  c'  =  116°, 
8  =  K57°  +  137°  +  116°)  =  155°, 

s-a'  =  98°,  s  -  b'  =  18°,  s  -  c'  =  39°. 

log  sin  (s  -  a')  =    9.9958 

log  sin  (s  -  b')  =    9.4900 

log  sin  (s  -  c')  =    9.7989 

log  esc  s  =    0.3741 

2)19.6588 

logA:=    9.8294 

log  k  =  9.8294  log  k  =  9.8294 

log  sin  (s  -  a')  =  9.9958  log  sin  (s  -  b')  =  9.4900 

log  tan  \  A'  =  9.8336  log  tan  \  E'  =  0.3394 

£  A'  =  34°  17',  I  B'  =  65°  24r, 

A'  =  68°  34',  B'  =  130°  48', 


X,  §80]       OBLIQUE    SPHERICAL   TRIANGLES  111 

log  Jc  =  9.8294 

log  sin  (s  -  cQ  =  9.7989 

log  tan  |  (7  =  0.0305 

I  a  =  47°  i', 

C'=94°2'. 
Therefore 

a  =  111°  26',  6  =  49°  12',  c  =  85°  58'. 

CHECK. 

log  sin  A  =  9.9236  log  sin  B  =  9.8338  log  sin  C  =  9.9537 

log  sin  a  =  9.9689  log  sin  b  =  9.8791  log  sin  c  =  9.9989 

9.9547  9.9547  9.9548 

NOTE.     Using  the  law  of  cosines  as  stated  in  (38),  .Art. 

71,  whence 

cos  B  cos  C  4-  cos  A 


cos  a  = 


sin  B  sin  (7 


and  proceeding  as  in  Art.  78,  the  following  formulae  may 
be  got : 

/     cos  (S  —  B)  cos  (S  —  C) 

or 


cos  (8  —  A)  cos  (8  —  B)  cos  (8  -  O) 
= 


where 


and  similar  formulse  for  cot  ^  b  and  cot  %  c.  These  formu- 
lae are  simple  and  convenient,  but  it  is  unnecessary  to  bur- 
den the  memory  with  them. 

80.  Cases  3  and  4,  two  sides  and  the  included  angle,  two 
angles  and  the  included  side,  are  solved  by  means  of  Napier's 
Analogies,  which  we  shall  proceed  to  derive. 

From  (48),  Art.  78,  we  may  write 

A  ^      B  * 

tan  ^-  tan  ^  = 


2          2      sin(s-a)  sin(s-6)' 


112  SPHERICAL   TRIGONOMETRY  [X,  §80 

or,  since  ,  2  _  sin  (s  -  a)  sin  (s  —  b)  sin  (s  —  c) 

sins 

sin  —    sin  — 

2  2  _  sin  (s  -  c) 

"2"'    ~^~       sins 
cos  —    cos  — 

2  2 

Whence, 

sin  -  sin  _ 
1  _      _2  _  2  _  1  _  sin  (s  -  c) 

cos^cos*'  sins 

2         2 
or, 

cos—  cos  --  sin  —sin  — 

22  22  _  sins-sin(s-c) 

eosfcosf  = 


That  is, 


cos  i  (A  +  B)  _  2  cos  |(2  s  —  c)  sin  |  c 

.4       J5  sins 

COS      C°S~ 


Whence,  since  2s  —  c  =  a-|-6-|-c  —  c  =  a  +  6, 

cos  J  (^  +  B)  _  2  cos  i  (a  4-  6)  sin  i  c  ,.~ 

^|^f=        S5T^' 

Also,  from  (a)  above, 

sin—  sin  — 
-,    ,      _2  _  2  _         sm(s-c) 

+= 


which  being  transformed  in  the  same  manner  gives 

cos  i  (A  —  B)  __  2  sin|(q  +  6)  cos  j-c 

"~Z        5~  :  sins 

cos  —  cos  — 

2         2 


X,  §80]       OBLIQUE    SPHERICAL   TRIANGLES  113 

Dividing  (ft)  by  (y)  we  have 

cosiQi  +  B)  =      tan|c  (49) 

cos|(4-5)      tan±(a+&) 

Again,  from  (48)  Art.  78  we  may  write 

tan  — 

_  2  _  sin  (s  —  6) 

^~  S~sin(«-a)' 

2 

sin  ^  cos  * 


2         2  _  sin  (s  — 


Whence, 


sin  —  cos  — 


2        2  sin(S-6) 


cos      sn 


sin  (s  —  a) 


sin  —  cos  —  ±  cos  —  sin  — 

2         2  _  2         2  _  sin  (s  —  b)  ±  sin  (s  —  a) 

^4    .     5  sin  (s  —  a) 

cos  —  sin  — 

2         2 

Using  the  upper  signs, 

sin  1  (A  +  B)  _  2sini  (2s  —  a—  6)  cos  -*-  (a  -  6) 
cos  |  sin  |  =  sin(s-tt) 

—  ^  gin  j  c  CQS  k  (&  —  &)  .  /sj\ 

sin  (s  —  a) 
Using  the  lower  signs, 

sin  %(A  -  B)  __  2  cos  |(2s  -  a  -  6)  sin  j-(a  -  6) 

sin  (8  -a) 


cos      sn 


—  2  cos  |  c  sin  ^  (a  —  b)  t  ,* 

sin  (s  —  a) 


114  SPHERICAL   TRIGONOMETRY  [X,  §80 

Dividing  (8)  by  (c), 

sin  -,  (A  +  B)  tan  £  c 

-;  =  -  -.  (50) 

sin  g  ( A  —  B)     tan  ^  (a  —  b) 

Applying  (49)  and  (50)  to  the  polar  triangles  we  obtain 

cos  i  (A1  +  B')  =        tan  1  c' 
cos  ^  (A'  —  B')      tan  \  (af  +  6') ' 

sin  ^  (A'  +  #')  =        tan  j-  c' 
sin  i  ( J.'  —  5')      tan  1  (a'  —  &') 

Remembering  that    A'  =  180°  -  a,    a'  =  180°  -  A,    etc. 
these  become 


cos  I  (a  +  ft)  cot  ^  C 

cos  I  (a  —  &)      tan  |  (A  -f- 


(51) 


££~  (^) 


The  formulae  (49),  (50),  (51),  and  (52),  called  Napier's 
Analogies  because  of  their  similarity  to  formula  (28)  of 
the  plane  trigonometry,  can  obviously  be  written  in  other 
forms  by  the  cyclical  interchange  of  the  letters. 

81.  Case  3.  Example.  Given  a  =  100°  30',  b  =  40°  20', 
(7=46°  40',  find  A,  B,  c.  Napier's  analogies  (51)  and  (52) 

may  be  written 

_.      cos  i  (a  -  6) 
; 


sm 


which  will  determine  A  and  B.     Then  to  find  c  we  may  use 
either  (49)  or  (50).     The  latter  may  be  written 


sm  %(A  —  B) 


X,  §81]       OBLIQUE   SPHERICAL  TRIANGLES  115 

We  have 
£(a  4-  b)  =  70°  25',  \  (a  -  6)  =  30°  5',  £  C  =  23°  20'. 

log  cos  £  (a  -  6)  =  9.9371  log  sin  £  (a  -  6)  =  9.7001 

log  cot  \  C  =  0.3652  log  cot  |  C  =  0.3652 

log  seel  (a  +  6)  =  0.4748  log  esc  %  (a  +  6)  =  0.0259 

log  tan  1(^1  +  £)=  0.7771  log  tan  ±(A-B)  =  0.0912 

i  (J.  +  B)  =  80°  31',  £  (A  -  B)  =  50°  59'. 

Whence  A  =  131°  30',  jB  =  29°  32'. 

log  sin  %(A  +  B)=  9.9940 
log  tan  i  (a  -6)  =9.7629 
log  esc  $(A-B)=  0.1096 
log  tan  i  c  =  9.8665 
|c  =  36°20f, 
c  =  72°40'. 

The  signs  are  all  plus  in  the  above  computation. 

Case  4.     Example.    Given  B  =  110°  40',  C  =  100°  36',  a  = 
76°  38',  find  6,  c,  A 

Napier's  analogies  (49)  and  (50)  may  be  written 

cos  iB-C  tan  1  a 


sin      5-  6*  tan 


which  will   determine  6  and  c.     To  find  A  either  (51)  or 
(52)  may  be  used.     The  latter  is 


sin  i  (6  —  c) 

Here  |  (5  +  C)  =  105°  38',  $(B  -  C)  =  5°  2',  1  a  =  38°  19'. 

log  cos  i(B—  0)  =  9.9983  +log  sin  1  (B  -  0)  =  8.9432 

+log  tan  i  a  =  9.8977  +log  tan  1  a  =  9.8977 

log  sec  ±(B+C)  =0.5695  +log  esc  |  (B  +  0)  =  0.0164 

-log  tan  £  (&  +  c)  =  0.4655  +log  tan  |  (6  -  c)  =  8.8573 

180°  -  $(b  +  c)  =  71°  6',  |(6  -  c)  =  4°  7'. 


116  SPHERICAL   TRIGONOMETRY  [X,  §81 

Whence  b  =113°  1',  c  =  104°  47'. 

log+  sin  i  (6  +  c)=  9.9759 
log+  tan  (£  -  O)  =  8.9449 
log+ csc  i  (6 -c)  =  1.1440 
log+  cot  i  ^1  =  0.0648 
i  ^  ==  40°  44', 
.4  =81°  28'. 

Note  that  the  algebraic  signs  are  not  all  plus,  and  that 
the  quadrant  in  which  the  angle  lies  is  determined  by  the 
sign  in  the  case  of  the  tangent,  cotangent,  or  cosine. 

82.  Cases  5  and  6,  two  sides  and  an  opposite  angle  or  two 
angles  and  an  opposite  side,  may  be  solved  by  the  law  of 
sines  together  with  Napier's  analogies.  Thus,  if  a,  b  and  A 
are  given,  we  may  write 

sin  b  sin  A 


sin  B  = 


sm  a 


which,  however,  does  not  determine  B  unambiguously,  since 
B  may  be  either  acute  or  obtuse.  In  this  case,  indeed, 
there  may  be  two  solutions,  one  solution,  or  none.  We 
know,  however,  that  if  two  sides  (or  angles)  of  a  spherical 
triangle  are  unequal  the  angles  (or  sides)  opposite  are  un- 
equal, and  the  greater  angle  (or  side)  lies  opposite  the 
greater  side  (or  angle).  These  theorems  enable  us  to  deter- 
mine which  values  of  the  angle  (or  side)  are  possible. 

Thus  ifb^a,  then  only  values  of  B  which  are  |  Qrea^er  \  ^ian 

A  are  possible;  both  values  of  B  may  be  so,  or  only  one 
value.  If  the  sine  of  B  is  greater  than  unity ;  that  is,  if  log 
sin  B  is  positive,  no  solution  is  possible.  These  same  con- 
siderations obviously  apply  to  case  6  also. 

Another  method  of  removing  the  ambiguity  of  Cases  5 
and  6  is  as  follows :  Two  angles  are  of  the  same  species 
when  they  are  both  acute  or  both  obtuse.  Also,  since  each 
side  and  angle  of  a  spherical  triangle  is  less  than  180°,  we 


X,  §82]       OBLIQUE    SPHERICAL   TRIANGLES  117 

see  that  $(A  +  B)  and  |  (a  +  b)  are  each  less  than  180°  ; 
while  %(A  —  B)  and  1  c  are  each  less  than  90°.  It  follows, 
in  Napier's  first  analogy, 

cos  i  (A  +  B)  _       tan^c 
cos  i  (A  -  B)  ~~  tan  |(a  +  b)  ' 

that  tan  ^  c  and  cos  ^  (  A  —  B)  are  both  positive.  Then 
cos  -i  (  A  +  B)  and  tan  \  (a  +  b)  must  have  the  same  algebraic 
sign,  and,  therefore,  1  (A  +  B)  and  |(a  +  6)  are  o/  £fte  same 
species.  Thus,  when  a  and  b  are  given  and  A  or  5  is  to  be 
found,  if  i(a  +  6)  ^  90°  m^s£  aZso  %(A+B)^  90°  ;  and 
the  values  of  A  or  5  must  be  so  chosen  as  to  satisfy  this 
condition. 

Having  thus  found  B,  say,  (whether  there  be  two  values 
or  only  one)  we  may  complete  the  solution  of  the  triangle 
by  the  use  of  Napier's  analogies. 

Example  1.      Given  a  =  46°  30',  b  =  30°  20',   B  =  36°  40', 

solve  the  triangle. 

We  have  .  sin  a  sin  B 

sin  A  =  — 

sm  b 

log  sin  a  =  9.8606 
log  sin  B  =  9.7761 
log  esc  b  =  0.2967 
log  sin  A  =  9.9334 
^1  =  59°  4'  or  120°  56'  =  ^'. 

Here  a  >  6,  and,  therefore,  must  A>  B.  This  is  true  of 
both  values  of  A  found,  so  that  there  are  two  possible  solu- 
tions of  the  triangle.  To  find  C  and  c  we  may  use  (52)  and 
(50). 

cot  C=  sin  JO  +  6)  tan  ±(A-B) 
2  sina-& 


,      c  = 


2  sin  |(^1  -J5) 


118  SPHERICAL   TRIGONOMETRY  [X,  §82 

We  have 

First  solution  Second  solution 

.1  (a  +  6)  =  38°  25,'  '  |(a  +  b)   =  38°  25', 

|(a-&)=8°6',  K«-&)   =8°  5', 

£(4  +  -B)  =  47°  52/  %(A'  +  B)  =  78°  48', 

l(A  -B)=  11°  12'.  i(^'  -  B)  =  42°  8'. 


log  sin  |(a  +  6)  =  9.7934     or    9.7934 

log  tan  $(A  -B)=  9.2966  9.9565 

logcsc^(a-&)  =0.8519  0.8519 

log  cot  -  =  9.9419     or     0.6018 

^=48°  49',    Y  =  14°  3'> 
(7  =97°  38'.     O'  =  28°6'. 

log  sin  %(A  +  B)  =  9.8701    or     9.9916 
log  tan  £(a  -  6)  =  9.1524  9.1524 

log  esc  %(A-  B)=  0.7117  0.1734 


log  tan  £=9.7342    or    9.3174 

-=28°  28',  -  =  11°  44', 

c=56°56'.  c'  =  23°38'. 

The  two  complete  solutions  are,  therefore, 

^1=59°  4',       or     120°  56', 
C  =  97°  38',  28°  6', 

c  =  56°56'.  23°  28'. 

Example  2.     Given  a  =  126°,  c  =  70°,  A  =  56°,  solve  the 
triangle. 

Using  the  formula 

sin  A  sin  c 


.     ^ 
sm  C 


sin  a 


X,  §82]       OBLIQUE    SPHERICAL   TRIANGLES  119 

we  have  log  sin  A  =  9.9186 

log  sin  c  =  9.9730 

log  esc  a  =  0.0920 

log  sin  C  =  9.9836 

C  =  74°  20'  or  105°  40'. 

But  since  a  >  c,  must  A  >  C.     Therefore,  there  is  no  sohtr 
tion.     Otherwise  thus  : 

£  (a  +  c)  =  98°,  i  (A  +  C)  =  65°  10'  or  80°  50',  which  are 
not  of  the  same  species. 

Example  3.     Given  A  =  84°,  C  =  19°,  a  =  28°,  solve  the 
triangle. 

Using  the  law  of  sines,  sin  c  =  —  ;  —  -  —  - 

sm^. 

log  sin  C=  9.5126 
log  sin  a  =  9.6716 
log  esc  ^4  =  0.0024 
log  sin  c  =  9.1866 

c  =  8°  50'  or  171°  10f. 

But   since    C  <  A,   must   c  <  a,   and   the   second  value    is 
impossible. 

To  find  b  use  (50). 

log  sin  ±(A  +  C)=  9.8935 

log  tan  i  (a  -  c)  =  9.2275 

log  esc  fr(^-Q  =  0.2698 


log  tan  i  6  =  9.3908 
|6  =  13°49r,  6  =  27°  38'. 

To  find  B  we  may  use  (52),  which  has  the  advantage  of 
giving  an  unambiguous  result,  or  the  law  of  sines.  Selecting 
the  latter  method  we  have 

log  sin  (7=  9.5126 
log  sin  b  =  9.6663 
log  esc  c  =  0.8137 
log  sin  B  =  9.9926 

B  =  79°  27'  or  100°  33'. 


120  SPHERICAL   TRIGONOMETRY  [X,  §82 

But  since  b  <  a,  must  B  <  A,  and  the  second  value  is  im- 
possible.    The  complete  solution  is,  therefore, 

c  =  8°  50',  b  =  27°  38',  B  =  79°  27'. 

83.   Delambre's  Analogies  or  Gauss's  Equations. 
Using  the  law  of  cosines  we  may  write 
cos  a  —  cos  b  cos  c 

COS  A  = : : — 

sin  b  sin  c 
Whence 

1  -  cos  A  =  2  sin'  1  A  =  (c°S  6  COS  c  +  sinfe  sin  c)~  cos  a 

sm  6  sin  c 

or, 

0    •  9A     cos(6  —  c)— cos  a_2  sini(a-h6  — c)sini(a— 5+c) 

L  Sin  —  = ; ; — ; — —    . *•* *  • 

2  sm  b  sin  c  -  sin  b  sm  c 

That  is,  • 

sm^l  _    /sin  (s  —  b)  sin  (s  —  c) 
2      ^  sin  6  sin  c 

_B  f* 

with  similar  formulae  for  sin—  and  sin  •— • 

£  2i 

In  the  same  manner,  by  adding  unity  to  each  side  of  the 
first  equation  of  this  article,  may  be  obtained  formulae  of 
which  the  type  is 

cos-  =  Jsingsin(g-«). 
2       ^      sin  b  sin  c 

From  these  obviously  follows 


.     A         B  _  sin(s-6)     sin  s  sin  (s  -  c) 

Dill  — -  OUO    — ; Al  • •       r 

2          2  sin  c        ^       sm  a  sm  b 

_  sin  (s  —  b)          C_  (a) 

sine  2 

In  the  same  way  we  obtain 


2  sm  c 


X,  §83]       OBLIQUE    SPHERICAL   TRIANGLES  121 

Adding  (a)  and  (/3), 

sin  4  cos  *  +  cos  ^  sin|  =  sin(.-a)+sin(.-6)cog  <7 
2222  sine  2 

Whence 

sin  4-  c  cos  4  (a  —  6)         (7 

sin  4  M  +  B)  =  -  -^  cos  -^ 

sm^ccos^c  2' 

or 


cos  i  c  2 

Similarly  may  be  obtained 

sA-£=™coa 


sin,  III 

cos    -  c  2 


cos  1(A-B)  =  —  TV*  T-  "/  sin  ^  IV 

sm  i  c  2 

which  are  the  analogies  or  equations  sought.  These  im- 
portant equations  may  be  conveniently  used  in  the  solution 
of  Cases  3  and  4  of  oblique  triangles. 

Example.     Given   a  =  132°  47',   b  =  59°  50',    C  =  56°  28', 
solve  the  triangle. 

We  have 
£(a  +  6)  =  96°  19',   i(a  -  6)  =  36°  29',   1(7  =  28°  14'. 

log  sin  i  (a  +  b)  =  9.9973,  log  sin  £  (a  -  6)  =  9.7742, 

log  cos  i  (a  +  6)  =  9.0414,  log  cos  £  (a  -  b)  =  9.9053, 

log  sin  -  =  9.6749,        log  cos  -  =  9.9450. 

2  2 

From  equations  II  and  IV, 

log  Jsin  i  c  sm$(A  -  B)  \  =  9.7742  +  9.9450  =  9.7192, 
log  { sin  |  c  cos  %(A-B)\=  9.9973  +  9.6749  =  9.6722. 


122  SPHERICAL   TRIGONOMETRY  [X,  §83 

Whence 

log  tan  $(A-B)  =  0.0470, 


From  I  and  III, 

log+  {  cos  i  c  sin  J  (A  +  JB)  j  =  9.9053  +  9.9450  =  9.8503, 
log-  \GOS  i  c  cos  £  (4  +  £)  }  =  9.0414  +  9.6749  =  8.7163. 
Whence  log-  tan  %(A  +  B)  =  1.1340. 

180°  -%(A  +  B)  =  85°  48',  |  (A  +  5)  =  94°  12'. 
Therefore,  .4  =  142°  18',   £  =  46°  6'. 

Also          log  sin  $(A  —  B)  =  9.8718. 


Therefore,  log  jsin  1  c  sin  %(A  —  B)\  =  9.7192 

_  log  sin  i(A-B)=  9.8718 

log  sin  i  c  =  9.8474 

|c  =  44°44',   c  =  89°28'. 

Possibility  of  Solution  by  Inspection  of  Data.  Before  at- 
tempting the  solution  of  a  spherical  triangle  it  may  be 
desirable  to  determine  whether  the  triangle  is  possible  with 
the  given  data. 

Case  1.  Given  the  three  sides.  The  triangle  is  always 
possible  if  the  sum  of  the  sides  is  less  than  360°,  and  if  no 
one  side  is  greater  than  the  sum  of  the  other  two.  This 
follows  at  once  from  well-known  geometrical  theorems. 

Case  2.  Given  the  three  angles.  This-  case  can  be  read- 
ily tested  by  the  criteria  of  Case  1  applied  to  the  polar 
triangle.  For  example,  the  triangle  A  ==  78°,  B  =  100°, 
C  =  160°  is  impossible  because  the  sides  of  the  polar  tri- 
angle, a'  =  102°,  V  =  80°,  c'  =  20°,  are  such  that  a'  >  6'  +  c'. 

Case  3,  given  two  sides  and  the  included  angle,  and  Case 
4,  two  angles  and  the  included  side,  are  always  possible. 

Cases  5  and  6  have  been  discussed  in  Art.  82. 


X,  §83]      OBLIQUE   SPHERICAL  TRIANGLES 


123 


EXAMPLES 

1. 

a 

— 

68°  25', 

14.    a 

- 

111°  20', 

27. 

A 

— 

132°, 

b 

=. 

71°  11', 

c 

= 

41°  30', 

B 

= 

139°  50', 

c 

= 

56°  57'. 

C 

= 

26°  10'. 

b 

= 

127°  10'. 

2. 

a 

= 

100°  8', 

15.   A 

— 

159°  1', 

28. 

A 

—  • 

79°, 

b 

= 

50°  2', 

C 

= 

36°, 

B 

= 

40°, 

c 

= 

60°  6'. 

a 

= 

9°  5'. 

c 

= 

108°. 

3. 

A 

= 

51°  59', 

16.  A 

= 

25°  20', 

29. 

a 

— 

40°, 

B 

= 

83°  55', 

C 

= 

153°  30', 

b 

= 

118°  21', 

C 

= 

58°  54'. 

a 

= 

73°  33'. 

A 

= 

29°  25'. 

4. 

A 

= 

142°  33', 

17.    B 

i= 

142°  30', 

30. 

C 

— 

148°, 

B 

= 

27°  53', 

C 

= 

71°  20', 

B 

= 

22°  20', 

C 

— 

32°  27'. 

c 

= 

39°  35'. 

c 

== 

136°. 

5. 

b 

= 

42°  10', 

18.   A 

=: 

110°  5', 

31. 

a 

— 

114°, 

c 

— 

96°  11', 

B 

= 

123°  20', 

c 

— 

148°, 

A 

= 

110°  5'. 

b 

= 

126°  55'. 

C 

= 

135°  V. 

6. 

a 

— 

146°, 

19.    a 

= 

59°  34', 

32. 

A 

— 

73°, 

c 

= 

69°  20', 

b 

= 

136°  11', 

B 

- 

81°  50', 

B 

= 

125°  10'. 

c 

= 

150°  2'. 

a 

= 

122°  47'. 

7. 

a 

- 

90°  50', 

20.    a 

= 

109°  24', 

33. 

B 

— 

61°  40', 

c 

= 

117°  50', 

c 

= 

81°  50', 

C 

= 

140°  15', 

B 

= 

120°  6'. 

A 

=3 

107°  40'. 

c 

— 

150°  25'. 

8. 

B 

•= 

41°  6', 

21.    a 

= 

99°  50', 

34. 

a 

= 

125°  16', 

C 

= 

122°  10', 

c 

= 

64°  10', 

b 

= 

151°  37', 

a 

= 

37°. 

A 

= 

96°  13'. 

c 

= 

75°  55'. 

9. 

A 

— 

135°, 

22.   A 

— 

35°  31', 

35. 

A 

— 

60°  40', 

C 

= 

50°  50', 

B 

= 

24°  43', 

C 

= 

105°, 

b 

= 

68°  50'. 

C 

= 

138°  25'. 

a 

= 

64°  30'. 

10. 

A 

= 

147°  30', 

23.    A 

- 

31°  20', 

36. 

a 

= 

55°  5', 

C 

= 

163°  10', 

C 

= 

122°  40', 

c 

= 

138°  5', 

b 

= 

76°  25'. 

b 

= 

40°  40'. 

A 

= 

42°  28'. 

11. 

a 

— 

29°  2', 

24.    a 

= 

120°  45', 

37. 

B 

•=. 

116°  6', 

b 

= 

14°  3', 

c 

= 

70°  25', 

C 

= 

73°  50', 

A 

= 

49°  5'. 

B 

= 

50°  16'. 

c 

= 

80°. 

12. 

b 

— 

98°, 

25.  A 

- 

120°  21', 

38. 

B 

= 

134°, 

c 

= 

36°, 

B 

= 

130°  22', 

C 

= 

51°, 

C 

= 

163°. 

C 

= 

140°  7'. 

a 

= 

70°  20'. 

13. 

a 

= 

132°, 

26.     c 

& 

109°  20', 

39. 

b 

— 

108°, 

b 

= 

56°, 

b 

= 

80°  20', 

c 

= 

40°, 

A 

= 

116°  18'. 

C  =  106°  50'. 

C  = 

:39°. 

124  SPHERICAL  TRIGONOMETRY  [X,  §83 

40.  a  =  58°  20',             42.  A  =  70°  5',  44.   A  =  115°, 
6  =  138°  5',  B  =  122°,  B  =  60°, 
c  =  116°  3'.  C  =  95°  4'.  C  =  135°. 

41.  a  =  61°,                  43.  flf =  60°,  45.    a  =  150°, 
c  =  97°,  6  =  120°,  6  =  160°, 

5  =  110°.  c  =  50°.  B  =  10°. 

46.    a  =  112°  30',  47.  ^1  =  20°  30', 

6  =  108°  40',  B  =  32°  30', 

c  =  140°10'.  0  =  124°  30'. 


CHAPTER  XI 
THE  EARTH  AS  A  SPHERE 

84.  Distances  on  the  Earth.     As  we  remarked  in  the  intro- 
ductory chapter,  plane   trigonometry  is   sufficient   for  the 
survey  of  small  areas.     For  larger  areas  and  in  navigation, 
except  in  the  most  refined  work,  the  Earth  is  treated  as  a 
sphere  and  we  make  use  of  the  principles  of  spherical  trigo- 
nometry already  enunciated. 

The  shortest  distance  between  two  points  on  the  Earth  is 
the  arc  of  a  great  circle  joining  them.  If  we  know  the 
number  of  degrees  in  that  arc  we  can  compute  its  length  by 
the  formula  s  =  xr  (Art.  47),  where  s  is  the  length  of  arc, 
x  the  angle  in  circular  measure,  and  r  the  radius  of  the 
sphere ;  in  this  case  3960  miles,  the  radius  of  the  Earth. 

Example.  Find  the  length  of  an  arc  of  26°  on  the  Earth's 
surface. 

26'= 

Therefore, 

s  =^  x  3960  =  12  .  ??  .  3960  =  1798  miles, 
yo  yo     7 

It  is  convenient  to  compute  and  remember  the  number  of 
miles  in  one  degree  of  arc  for  the  purpose  of  finding  lengths 
of  arcs. 

s  =  -?-  x  3960  =  69.1  miles,  approximately. 

JLoU 

85.  Position  and  direction.     The  position  of  a  point  on  the 
Earth  is  determined  by  its  latitude  and  longitude;  that  is, 
by  the  number  of  degrees  the  point  lies  north  or  south  of 
the  equator,  and  the  number  of  degrees  east  or  west  of  a 

125 


126  SPHERICAL  TRIGONOMETRY          [XI,  §85 

great  circle,  through  the  Earth's  axis,  chosen  as  a  reference 
line.  We  shall  use  the  great  circle,  or  meridian,  through 
Greenwich. 

A  point  moving  along  a  great  circle  of  the  Earth,  unless 
that  circle  be  a  meridian  or  the  equator,  is  constantly 
changing  its  direction,  or  course.  For  example,  Fig.  36,  at 
A  the  compass  points  north  along  AN,  and  a  ship  at  A  is 


FIG.  36. 

sailing,  say,  due  west.  When  the  ship  has  reached  B  the 
compass  points  north  along  BN  and  the  ship  is  sailing  west 
30°  south.  On  the  other  hand  if  a  ship  sailed  constantly  on 
a  course,  say,  west  30°  south  it  would  move  around  the 
Earth  in  a  spiral  approaching  continually  nearer  to  the 
South  Pole. 

86.  Bearings.  To  illustrate  the  use  of  spherical  trigo- 
nometry in  determining  positions,  directions,  and  distances 
on  the  Earth's  surface,  consider,  Fig.  36,  a  ship  sailing 
from  C  to  A  along  the  great  circle  CBA.  The  lines  NC, 
NB,  NAy  and  NG  are  meridians,  the  last  being  the  meridian 
of  Greenwich.  Suppose  the  latitude  and  longitude  of  C 
are  44°  40'  N.,  63°  35'  W. ;  of  A,  53°  24'  K,  3°  4'  W.  The 
longitude  of  G,  obviously,  is  0°.  The  positions  of  the  points 
C  and  A  being  thus  known,  let  us  find  the  directions  (called 
of  A  from  C  and  of  C  from  A,  and  the  distance 


XI,  §86]  THE   EARTH   AS   A   SPHERE  127 

from  C  to  A.     From  the  meaning  of  longitude  we  have 

3°4:',   GNC  =  63°35',  whence   a  =  ANC  =  60°  31'. 
Also,  by  the  meaning  of  latitude, 
=  go0-  53°  24'  =  36°  36'  ;  CN=  90°  -  44°  40'  =  45°  20'. 


We  therefore  have,  in  the  spherical  triangle  CNA,  CN  = 
a  =  45°  20',  AN=  c  =  36°  36'  and  the  included  angle  a  = 
60°  31',  which  is  case  3  in  the  solution  of  spherical  triangles. 
The  data  : 

£  (a  +  c)  =  40°  58',  i  (a  -  c)  =  4°  22',  |  a  =  30°  15.5'. 

log  cos  i  (a  -  c)  =  9.9987  log  sin  1  (a  -  c)  =  8.8816 

log  cot  -  =  0.2340  log  cot  *  =  0.2340 

2  2i 

log  seel  (a  +  c)  =0.1220  log  esc  |(a  +  c)  =  0.1834 

log  tan  $(A+C)=  0.3547  log  tan  1  (A  -  C)  =  9.2990 

£  (A  +  C)  =  66°  10'  i  (A  -  C)  =  11°  16'. 

Whence           A  =  77°  26'.  C  =  54°  54'. 

Therefore  the  bearings  of  C  from  A  are  N.  77°  26'  W.  ;  of 
A  from  (7,  K  54°  54'  E. 

To  find  the  side  CA  =  x  we  have 

log  sin  i  (>1+  C)  =  9.9613 

log  tan  i  (a  -  c)  =  8.8829 

log  esc  %(A  -  C)  =  0.7092 


log  tan  -=9.5534 

£  a  =19°  47.5',  x  =  39°  35'. 

Therefore  length  CA  =  39.6°  x  69.1  miles  =  2736  miles. 

If  only  the  distance  sailed  is  required  it  is  simpler  to  use 
the  law  of  cosines.  Thus,  cos  x  =  cos  c  cos  a  -f  sin  c  sin  a 
cos  a. 


128  SPHERICAL   TRIGONOMETRY          [XI,  §86 

log  cos  c  =  9.9046  log  sin  c  =  9.7754 

log  cos  a  =  9.8469  log  sin  a  =  9.8520 

9.7515  log  cos  «  =  9.6921 
number  =    .5643  9.3195 

_  .2087  number  =    .2087 
cos  x  =  0.7730, 
x  =  39°  23' 

and  the  distance  =  39.4°  x  69.1  =  2723  miles. 

87.  The  course  of  the  ship  at  C  would  be  N.  54°  54'  E.  To 
show  how  the  ship's  course  changes  as  it  sails  along  CA  let 
us  find  the  course  as  the  ship  crosses  the  meridian  38°  W.  at 
the  point  B,  Fig.  35.  In  the  triangle  NCB  we  have  b=CN 
=  45°  20',  C  =  NCB  =  54°  54',  N=  CNB  =  63°  35'  -  38° 
=  25°  35'  ;  that  is,  two  angles  and  the  included  side. 

i(C+  N)  =  40°  14.5',  ±(C-N)  =  14°  39.5',  £  b  =  22°  40'. 

log  cos  ±(C-N)  =  9.9856         log  sin  |  (<7  -  N)  =  9.4033 
log  tan  i  b  =  9.6208  log  tan  |  b  =  9.6208 

=  0.1173          log  esc  $(C  +  N)  =  0.1897 


log  tan  £  (c  +  n)  =  9.7237  log  tan  |  (c  -  n)  =  9.2138 

£  (c  +  »)  =  27°  53.5'  £  (c  -  n)  =  9°  17.5'. 

Whence  c  =  BN=37°  11',  n  =  CB  =  18°  36'. 

The  latitude  of  £  is  90°  -  BN  =  52°  49'  N.,  and  the  dis- 
tance sailed  is 

CB  =  18.6°  x  69.1  miles  =  1285  miles. 
To  find  the  angle  B  =  CBN  we  have 

log  sin  i  (c  +  n)  =  9.6700 

log  tan  i  (C  -  N)  =  9.4176 

log  esc  i  (c  -  n)  =  0.7919 

log  cot  %  B  =  9.8795 

\  B  =  52°  51',  B  =  105°  42',  and  NBA  =  180°  -  B  =  74°  18'. 

Therefore  the  ship's  course  at  J5  (the  bearing  of  A  from 
B)  is  N.  74°  18'  E. 


XI,  §88]  THE    EARTH   AS  A   SPHERE  129 

88.  The  Area  of  a  Spherical  Triangle  may  be  found  as  fol- 
lows :  The  theorem  has  been  proved  that  the  area  of  a 
spherical  triangle  is  equal  to  its  spherical  excess  (the  excess 
of  the  sum  of  its  angles  over  two  right  angles)  times  the 
area  of  the  tri-rectangular  triangle;  it  being  understood 
that  the  right  angle  is  the  unit  of  angles.  Thus,  using  A  to 
represent  the  area  of  a  triangle  whose  angles  (in  degrees) 
are  A,  B,  and  C  ;  and  noting  that  the  tri-rectangular  triangle 
is  one  eighth  of  the  surface  of  the  sphere  ;  we  have 


A  +  B+C-ISQ0   4  irr*  =  (A  +  B  +  C  -  180)in* 
90°  8  180° 

Example.     Given  A  =  105°,  B  =  80°,  C  =  95°,  and  taking 
r  =  3960  miles,  the  radius  of  the  earth, 

A  =  (105°  +  80°  +  95"  -  180°)      /3960N2  =  5?r(3960)2 
180°  9 

log  5  =  0.6990 

log  TT  =  0.4971 
2  log  r  =  7.1954 
colog  9  =  9.0458  -  10 

log  A  =  7.4373 
and  A  =  27,370,000  square  miles. 

TABLE   OF  LATITUDE   AND  LONGITUDE 
Baltimore  39°  17'  K,    76°  37'  W. 


Boston 

42°  21'  N., 

71°  4'  W. 

Chicago 

41°  53'*K, 

87°  38'  W. 

Greenwich 

51°  29'  N., 

0°W. 

Honolulu 

21°  18'  N., 

157°  55'  W. 

Liverpool 

53°  24'  K., 

3°  4'  W. 

New  York 

40°  43'  K, 

74°  W. 

Pernambuco 

8°S., 

34°  W. 

Rio  de  Janeiro 

22°  54'  S., 

43°  10'  W. 

San  Francisco 

37°  48'  K, 

122°  24'  W. 

Washington 

38°  54'  N., 

77°  3'  W. 

K 

130  SPHERICAL   TRIGONOMETRY          [XI,  §88 


EXAMPLES 

In  the  following  problems  assume  that  one  can  travel  directly 
along  the  arc  of  a  great  circle  between  the  points  named. 

1.  A  ship  sails  from  Baltimore  to  Boston.     Find  the  course  of  the 
ship  as  she  leaves  Baltimore,  her  course  as  she  crosses  the  meridian  of 
New  York,  and  the  entire  distance  she  sails.     What  are  the  bearings 
of  Baltimore  from  Boston,  and  of  Boston  from  Baltimore  ? 

2.  Find  the  course  at  Liverpool,  the  course  at  55°  W.,  and  the 
total  distance  sailed  by  a  ship  going  from  Liverpool  to  New  York. 
What  are  the  bearings  of  these  cities  from  each  other  ? 

3.  A  ship  sails  from  Baltimore  to  Rio  de  Janeiro.     She  sails  first 
to  a  point  off  Pernambuco  in  latitude  8°  S.,  longitude  34°  W.,  and 
from  there  to  Rio.     How  far  does  she  sail,  and  what  is  her  course  off 
Pernambuco  ? 

NOTB.  In  the  Southern  Hemisphere  latitudes  are  taken  as  algebra- 
ically negative.  Use  the  north-polar  distances  of  places  as  sides  in 
solving  triangles. 

4.  In  problem  3  what  course  will  the  ship  be  sailing  after  she  has 
gone  1000  miles  ?     What  will  be  her  latitude  and  longitude  at  that 
point  ? 

5.  How  far  is  the  Washington  Observatory  from  the  Greenwich 
Observatory  ?     What  are  the  bearings  of  the  two  observatories  from 
each  other  ? 

6.  A  ship  sails  from  Boston  on  a  course  East  12°  South.    At  what 
distance  would  she  be  sailing  due  East  ?    What  are  her  latitude  and 
longitude  at  that  instant  ? 

7.  A  ship  sails  northwest  from  San  Francisco.    What  would  be 
the  highest  latitude  she  would  reach?    What  would  be  the  ship's 
longitude  at  that  instant  ? 

8.  Find  the  number  of  square  miles  in  the  triangle  whose  vertices 
are  Baltimore,  Boston,  and  Chicago. 

9.  A  ship  sails  from  Honolulu  to  San  Francisco.    Find  the  entire 
distance  sailed,  and  the  course  of  the  ship  when  she  has  gone  halfway. 

10.  An  aeroplane  sails  from  Washington  to  Chicago  along  a  great 
circle  arc  one  mile  above  the  surface  of  the  Earth.  In  what  time  is 
the  flight  made  at  a  rate  of  75  miles  per  hour  ? 


XI,  §88]  THE    EARTH   AS   A   SPHERE  131 

11.  Find  the  number  of  square  miles  in  the  triangle  whose  vertices 
are  Baltimore,  New  York,  and  Chicago. 

12.  Find  the  face  and  edge  angles  of  a  regular  triangular  pyramid. 

13.  What  is  the  latitude  of  three  points  on  the  Earth  equally  dis- 
tant from  each  other  and  from  the  North  pole  ? 

14.  Each  face  of  a  triangular  pyramid  is  a  triangle  whose  sides  are 
3,  4,  and  5  respectively.    Find  the  face  and  edge  angles  of  the  pyramid. 


ANSWERS 


1.    6  =  340 

c  =  422 

4.    6  =  478 

a  =154 

7.   70.7ft. 

12.   99.5ft. 


8.    1.912 
17.    100 


CHAPTER   III 

» 

Art.  28 

2.   A  =  43°  5 
c  =  54.9 
5.     a  =  713 
c  =  823 
8.   71.2ft.       9.   60°  10' 


3.   ^  =  51°  47' 
b  =  .433 

6.     b  =  96.4 
c  =  232 
10.  260ft.        11.    212ft. 


13.   43°  36'        14.   Heights  equal. 

Art.  35 

9.    -  .874  14.   c2  15.   10°+*  16.   e5 

20.   0  21.   0  22.    1  23.   Impossible. 


2(logea-loge6) 
a  —  b 


27.   0         29.   .6931 


30.  1.099  31.  1.386  32.  .2312  33.  1.029  34.  -.3088 
35.4.408  36.238.2  37.  -  358300  38.  .07212  39.  Impossible. 
40.  -312.1  41.  -1747  42.  57090  43.  .00003162 

44.  .03728  45.  100  46.  3.241  47.  .001347  48.  1142 
49.  2448 

SOLUTIONS  OF  RIGHT  TRIANGLES 


1.     a  =  .04691         2.      a  =  2316 
c  =  .05151  b  =  3402 

IT  =  .0004988  K  =  3,941,000 

5.      a  =  578.8 
c  =  2491 
K=  701,000 
8.      a  =  .6441 
c  =  .6503 


b  =  48.04 
#=1859 
7.  A  =  63°  48' 

c  =  .4694 
#=.04364 


#=.02879 
133 


3.     b  =  24850 
c  =  36100 
K=  325, 400,000 
6.     a  =  .00883 
b  =  .003607 
K=  .00001593 
9.    J.  =  43°24' 
6  =  .8966 
#=.3801 


134  ANSWERS 

10.     a  =  .  003845     11.     6  =  5091  12.     6  =  99.43 

b  =  .006723               c  =  5268  c  =  156.8 

K  =  .00001293          K  =  3,444,000  #  =  6030 

13.    A  -  79°  28'          14.     6  =  63,840  15.     a  =  .000005737 

a  =  842                         c  =  92,280  c  =  .00002118 
K  =  65,900                   K  =  2,128,000,000       K  =  .0000000000585 

16.     a  =  .0003899       17.   A  =  27°  17'  18.     a  =  18.59 

6  =  .0006772                a  =  4.252  6  =  30.51 

1T  =  .000,000,1321       K=  17.53  JT=  283.7 

19.     6  =  24,540           20.  ^  =  43°  45' 

c  =  30,010  c  =  5280 

#  =  211 ,900,000  K  -  6,970,000 

21.   First  steeper  by  54'.  22.   24.7  mi:  and  29.5  mi. 

23.   Team  by  15  seconds.  24.   3°  25'      25.   648ft.      26.    14.7  in. 

27.  9  hr.  28  min.  A.M.  or  2  hr.  32  min.  P.M. 

28.  Reduced  by  10.1  ft.  29.    Buoy  farther  by  1133  ft. 

30.  Increase  in  altitude  251.4  ft.        31.    8°  45'.       32.    1  ft.  shorter. 

33.  1575  mi.         34.    57°  43'  N.  or  S.          35.    N:  58°  15'  E.     15  mi. 

36.  E.  62°  46'  N.     7.29  mi.  per  hour. 

37.  112.5  mi.  38.   E.  80°  N.  or  S.     4.33  mi.         39.   127.9  mm. 
40.  155.1ft.           41.    74.17yd. 


±V2 
2 


CHAPTER  IV 

Art.  40 

5. 
9. 

9  ±4x/7 
20 

±V2 

a     1  ±  2V30 

12 

10     ±7V^ 

10 

10 

Art.  41 


4  5     J=9V3  j=  8\/2  6     ±32\/2±9\/15 

5  7 

7.    One  value,  45°  8.    ±  -27S  0  9.  Two  values,  ±  90° 


Art.  44 


5.    ±±1  6.    ±(3±2V2) 

V6 


ANSWERS 


135 


18. 

23. 
25. 

CHAPTER   V 
Art.  47 

Minute  hand,  TT       19.    —       20.    —       21.   —       22.    14  T 
6                   20                   9 
Hour  hand,  — 

—  radians  per  second.             24.    191  revolutions  per  minute. 
3 

6934  mi.          26.    1978V2  mi.          27.   .035  in.          28.   6.7ft. 

Art.  49 

17. 

V2  +  V6 
4 

18     V6-V2              19 

512               20     0 

-LO.                                                                   -LC7.       • 

4 

3 

CHAPTER   VI 

1. 

6  =  986 
c  -  544.3 
#  =  193,000 

2.    A  =  19°                      3. 
0  =  52° 
c  =  8.19 
#  =  13.1 

A  =  18°  39' 
B  =  26°  52' 
c  =  673.9 
#=45,990 

4. 

B  =  59°  18' 
C  =  71°  36' 
#=1705 

5.   A  =  100°  35'  or  10°  21' 
C  =  44°53'        135°  7' 
a  =  67.02           12.25 
#  =  914              167.1 

6. 

a  =  6184 
6  =  2937 
#=7,510,000 

7.    5  =  49°  8'                    8. 
C  =  59°  19' 
a  =  70.48 
#=1703 

A  =  37°  58' 
B  =  66'  42' 
a  =  179.9 
#=23,370 

9. 

a  =  6.64 
c  =  3.95 
#=12.21 

10.    £  =  23°  29'                11. 

C  =  22°  57' 
b  =  3024 
#=3,243,000 

C  =  61°28' 
c  =  .4592 
#=.0802 

12. 

A  =  94°  16' 
B  =  54°  36' 
C  =  31°  8' 
#=  .0002699 

13.   No  solution.               14. 

b  =  .01292 
c  =  .002861 
#=.00000826 

15. 

A  =  26°  19' 
C  =  109°  6' 
c  =  67.14 
#=742 

16.    b  =  .0185                   17. 
A  =  54°  40' 
C  =  94°  13' 
#  =  .000002694 

5  =  90° 
c  =  59.39 
#=955 

136  ANSWERS 

18.   J.  =  46°23'      or  133°  37'  19.    #  =  22°  37' 

C  =  102°  30'  15°  16'  C  =  127°  28' 

c  =  8730  2354  a  =  .5593 

K  =  14,600,000  3,938,000  K  =  .0958 

20.    A  =134°  22'  21.   4  =  57°  41' 
B  =  18°  42'  C  =  38°  49' 

C  -  26°  56'  a  =  .02461 

K=  4.622  K=.  0002232 

22.  33,695  sq.  ft.            23.    15  ft.           24.    35.6  ft.          25.    428  ft. 

26.  Width  74.6  ft.  ;  height  above  stream,  14  ft.           27.   472  ft. 

28.  17.1ft.                   29.    109ft.                    30.   61.93  and  58.81  ft. 

31.  A  =  61°  43'  £  =  80°  7'              C  =  38°  10'              c  =  5.20 

32.  B  =  53°  26'  a  =  46.45  ft.          c  =  52.48  ft. 

33.  11,320  and  7082.  34.   3997  sq.  ft.           35.   A  =  7°  5'  ;  no. 
36.  21.7  mi.                          37.   9°  17'  with  the  surface  of  the  water. 
38.  33.5  ft.          39.   40  ft.          40.    Second  yacht  by  1  min.  12  sec. 

41.  At  82°  33'  with  shore  on  the  side  towards  the  60°  angle. 

42.  A-C-B\)j  $560.  43.   AC  =152.1  ft.     BC  =  319.4  ft. 
44.   441ft.  45.    336.9ft. 

CHAPTER   VII 

1.    riTT  2.  WTT  ±  |  3.    2  nir  ±  ^ 


4.  2  wr  +  cos-'  5.  2  iw  ±  6.  rnr 

37  •" 


2  T17T  +  COS"1  £  2  T17T  ±  £• 

o 


7.   2nT-  8.  iiT+  9. 


2n?r      TT  riTT      TT 

36  28 


10.   2  riTr  ±  coB-=v  11.  nv.  12.  ?ITT  ± 

2nir 


ANSWERS 


137 


13. 


16. 


mr  .    TT 
~3~      12* 


14.    — 
4 


»•  T+^ 


15. 


18.   TIT  +  tan-i  f 


19. 
21. 


mr      TT 

2       8 


4  20.  2ri7r±^,    riTTitan-i^ 

4  ' 

22.  0  =  2  mr  ±  -  23.  0  =  n?r,  2  nir  ± 

2 


^1±JL£  r  =  2 

'    2 

2       4 


25.  x  = 

y  = 


=  0,    * 


,  n  even 


av/3 


26.    *  =       + 


ir*±=^P»  nodd 
5 


27.   fl  = 


r  =  ± 


28.  x  =       +   -» 


5vf 

2 

3 


[.  e=  —  .  r=0,  n  even 
2 


32. 

35. 
39. 

47. 

6  =  mr  +  - 
4 

r  =  ±2>/2 

33.  0  = 

7*  =: 

36.   ±4 
40.   ±1 

44     "" 
2' 

48.   0 

=  ±i         34.  «V 

±  2V2 
v/c             07      I    v  2 

'i  _  52  +  6x/r 

38.  1  or 

'7£ 

42        1 

i 

14 

0,   ±V3 

ka 
-y-i 

41      1  T/3    ±V 

t 

or  £ 

2 
2"~ 

'      3 

-^                45.  0. 
6 

_T        7T                     4q 

~2'    6             49<  X 

46 

=  isin~1(  —  ; 

2/  = 


V2V5-2 


138 

Kf\ 

ANSWERS  , 

0       ST                                 KI       "• 

50           ,2-8>/6 

O\J. 

z: 

in37r 
8 

10 

10 

CHAPTER  IX 

1. 

B 

— 

58°  30' 

6 

=  35°  59' 

c  — 

43°  33' 

121°  30' 

144°  1' 

136°  27' 

2. 

A 

= 

30°  53' 

a 

=  30°  13' 

c  = 

78°  35' 

149°  7' 

149°  47' 

101°  25' 

3. 

A 

— 

48°  11' 

a 

=  44°  29' 

c  = 

109°  52' 

131°  49' 

135°  31' 

70°  8' 

4. 

A 

— 

83°  39' 

5.   A 

=  159°  39' 

6.   A  = 

147°  34' 

B 

= 

127°  20' 

B 

=  104°  14' 

B  = 

66°  3' 

a 

= 

82°  1' 

b 

=  136° 

a  = 

157°  26' 

7. 

A 

— 

27° 

8.   A 

=  81°  29' 

9.   A  = 

139°  5' 

B 

— 

73° 

B 

=  131°  50' 

Tt    

110°  57' 

c 

— 

53°  8' 

c 

=  97°  42' 

c  = 

63°  47' 

10. 

a 

=3 

49°  26' 

11.    a 

=  147°  37' 

12.    a  = 

53°  45' 

b 

— 

43°  58' 

b 

=  136°  32' 

b  = 

153°  17' 

c 

= 

62°  5' 

c 

=  52°  11' 

c  = 

121°  53' 

13. 

a 

3 

7°  59' 

14.    a 

=  49°  11' 

15.    o  = 

129°  30' 

b 

- 

21°  58' 

b 

=  100° 

6  = 

166°  50' 

B 

= 

70°  59' 

A 

=  49°  37' 

B  = 

163°  8' 

16. 

A 

= 

30°  47' 

17.   A 

=  126°  53' 

18.    a  = 

141°  47' 

b 

— 

11°  36' 

b 

=  32°  29' 

c  = 

140°  37' 

c 

=: 

13°  26' 

c 

=  133°  18' 

B  = 

16°  25' 

19. 

B 

— 

10°  23' 

6 

=  6°  16' 

c  = 

142°  41' 

169°  37' 

173°  44' 

37°  19' 

20. 

A 

— 

170°  50' 

21.   A 

=  25°  5' 

22.    a  = 

66°  12' 

B 

— 

84°  45' 

B 

=  114°  38' 

b  = 

146°  25' 

b 

= 

55°  1' 

c 

=  168°  23' 

c  = 

109°  39' 

23. 

a 

— 

142°  40' 

24.  A 

=  28°  19' 

b 

— 

78°  7' 

b 

=  110°  59' 

A 

= 

142°  3' 

c 

=  108°  39' 

25. 

b 

— 

33°  37' 

c 

=  101° 

B  = 

34°  20' 

146°  23' 

79° 

145°  40' 

26. 

A 

— 

74°  12' 

27.   A 

=  68°  11' 

28.    a  = 

161°  32' 

B 

— 

157°  47' 

B 

=  39°  43' 

b  = 

129°  57' 

a 

— 

43°  57' 

c 

=  61°  11' 

c  = 

52°  28' 

ANSWERS 


139 


31. 
32. 

35. 

29.    a  =  168°  30'                         30. 
b  =  130°  29' 
B  =  99°  40' 

b  =  20°  23'                     B  =  10°  35' 
159°  37'                            169°  25' 

JL  =  141°32'             33.    ^  =  163°  16' 
C  =  111°  46'                     B  =  19°  55' 
b  =  108°  29'                     a  =  138°  36' 

B  =  23°  26'                       C  =  30°  51' 
156°  34'                            149°  9' 

36.    B  =  12°  7' 
C  =  139°  5' 

a  =  75°  40' 

A  =  124°  32' 

6  =  16°  48' 
c  =  151°  57' 

C  = 

34.    a  = 
b  = 
B  = 

b  = 

148°  10' 
31°  50' 

124°  38' 
46°  49' 
33°  43' 

50°  50' 
129°  10' 

CHAPTER   X 

1. 

4..= 

76° 

2.   A 

— 

138°  18' 

3.    <z  = 

38°  2' 

B  = 

81° 

B 

- 

31°  12' 

6  = 

51°  2' 

C  = 

61° 

C 

= 

35°  52' 

c  = 

42°  2' 

4. 

a  = 

101° 

2' 

5.    B 

— 

41°  32' 

6.   A  = 

145°  23' 

b  = 

49° 

C 

- 

79°  2' 

C  = 

108°  3' 

c  = 

60° 

a 

= 

108°  10' 

b  = 

126°  24' 

7. 

A  = 

105° 

57' 

8.    b 

= 

47°  49' 

9.    a  = 

120°  25' 

C  = 

121° 

45' 

c 

- 

72°  37' 

c  = 

71°  1' 

b  = 

115° 

56' 

A 

=3 

32°  16' 

B  = 

49°  52' 

10. 

a  = 

124° 

57' 

11.    B 

— 

22°  13' 

12.    No  solution 

c  = 

153° 

47' 

C 

— 

112°  8' 

B  = 

140° 

26' 

c 

= 

36°  30' 

13. 

T>  

90° 

14.   A  = 

36°  42'       B 

=  160°  32'     b 

=  148°  44' 

c  = 

138°  32' 

143°  18' 

38°  52' 

78° 

c  = 

146° 

42' 

15. 

No  solution 

16.     c 

— 

90° 

17.    a  = 

138°  34' 

b 

= 

18°  15' 

b  = 

155°  50' 

B 

= 

8°  2' 

A  = 

100°  16' 

18. 

a  = 

63°  59' 

c 

= 

156°  10' 

c  = 

155°  2' 

116° 

1' 

72°  54' 

87°  36' 

19. 

A  = 

110° 

4' 

20.    6 

_ 

115°  19' 

21.    B  = 

96°  16' 

B  = 

131° 

2' 

B 

— 

114°  2' 

C  = 

65°  14' 

C  = 

147° 

2' 

C 

— 

90° 

b  = 

99°  52' 

140 


ANSWERS 


22. 

a 

— 

61°  2' 

23.    B  = 

37°  30' 

24.     6  = 

69°  46' 

b 

— 

39°  2' 

a  = 

33°  49' 

A  = 

67°  37' 

c 

= 

92°  2' 

c  = 

64°  15' 

G  = 

25°  17' 

25. 

a 

= 

90°  58' 

26.    a  = 

120°  29' 

b 

— 

118° 

A  = 

119°  3' 

c 

= 

132°  2' 

B  = 

90° 

27. 

a 

- 

66°  40' 

c  = 

160°  54' 

C  = 

164°  38' 

113°  20' 

102°  2' 

127°  42' 

28. 

C 

= 

109°  58' 

a 

— 

96°  42' 

b 

= 

40°  34' 

29. 

B 

= 

42°  15' 

C  = 

160°  10' 

c  = 

153°  38' 

137°  45' 

49°  50' 

89°  56' 

30. 

A 

= 

44°  54' 

31.   No  solution. 

32.  No  solution 

a 

— 

112°  14' 

b 

= 

29°  53' 

33. 

A 

= 

89°  36' 

a  = 

138°  54' 

6  = 

42°  48' 

26°  48' 

25°  28' 

137°  12' 

34. 

A 

= 

142° 

35.    B  = 

58°  6' 

36.     6  = 

96°  24' 

B 

= 

159° 

6  = 

61°  31' 

B  = 

54°  54' 

C 

= 

133° 

c  = 

90° 

C  = 

146°  38' 

37. 

A 

— 

72°  54' 

38.   A  = 

51°  16' 

a 

— 

78°  30' 

b  = 

119°  47' 

b 

= 

112°  57' 

c  = 

69°  41' 

39. 

a 

— 

129°  44' 

A  = 

131°  8' 

B  = 

68°  36' 

95°  40' 

76°  58' 

111°  24' 

40. 

A 

= 

70°  44' 

41.     6  = 

110°  50' 

42.    a  = 

62°  42'. 

B 

- 

132°  12' 

A  = 

61°  35' 

6  = 

126°  44' 

C 

— 

94°  54' 

C  = 

86°  27' 

c  = 

109°  42' 

2. 


CHAPTER   XI 

Course  at  Baltimore  :  E.  37°  55'  N. 

Course  at  meridian  of  New  York  :  E.  36°  14'  N. 

Bearings  Boston  from  Baltimore  :  E.  37°  55'  N. 

Bearings  Baltimore  from  Boston  :  W.  34°  17'  S. 

Distance  sailed  :  359  mi. 

Course  at  Liverpool :  W.  14°  51'  N. 

Course  at  55°  W.  :  W.  26°  56'  S. 

Bearings  New  York  from  Liverpool :  W.  14°  51'  N. 

Bearings  Liverpool  from  New  York  :  E.  40°  31'  N. 

Distance  sailed  :  3303  mi. 


ANSWERS  141 

3.  Course  at  Pernambuco  :  S.  36°  35'  E.  arrives. 
Course  at  Pernambuco  :  S.  29°  33'  W.  departs. 
Distance  sailed :  5141  mi. 

4.  Course  :  S.  42°  32'  E. 
Position  :  29°  12'  N.,  64°  1'  W. 

5.  Bearings  Greenwich  from  Washington  :  E.  40°  41'  N. 
Bearings  Washington  from  Greenwich  :  W.  18°  33'  N. 
Distance :  3669  mi. 

6.  Distance  :  11,550  mi. 
Position  :  43°  42'  S.,  91°  25'  E. 

7.  Position  :  66°  2'  N.,  153°  54'  W. 

8.  117,700  sq.  mi. 

9.  Distance  :  2398  mi. 
Course  :  E.  29°  6'  N. 

10.    8  hours,  nearly.  11.    35,580  sq.  mi. 

12.  Face  angle,  60°  ;  edge  angle,  70°  32'. 

13.  19°28'S. 

14.  Face  :  36°  52',  53°  8',  90°. 
Edge  :  180°,  0°,  0°. 


Printed  in  the  United  States  of  America. 


HE  following  pages  contain  advertisements  of  a 
few  of  the  Macmillan  books  on  kindred  subjects. 


Descriptive  Geometiy 


BY  ERVIN   KENISON 
Associate  Professor  of  Drawing  and  Descriptive  Geometry 

AND 
HARRY  C.   BRADLEY 

Assistant  Professor  of  Drawing  and  Descriptive  Geometry  in  the 
Massachusetts  Institute  of  Technology 

Edited  by  Professor  E.  R.  HEDRICK 


The  point  of  view  maintained  in  this  text  is  that  of  a 
draftsman.  Mathematical  formulae  and  analytic  computa- 
tions have  been  almost  entirely  suppressed.  The  method  of 
attack  throughout  is  intended  to  be  that  which  shall  most 
clearly  present  the  actual  conditions  in  space.  Wherever 
experience  has  shown  that  a  simple  plan  and  elevation  are 
not  sufficient  for  this  purpose,  additional  views  or  pro- 
jections have  been  introduced  freely,  corresponding  to  the 
actual  drafting  practice  of  making  as  many  side  views  or 
cross  sections  as  may  be  needed. 

The  amount  of  ground  covered  by  this  book  is  that  which 
is  considered  sufficient  to  enable  the  student  to  begin  the 
study  of  the  technical  drawings  of  any  line  of  engineering 
or  architecture.  It  is  not  intended  to  be  a  complete  treatise 
on  descriptive  geometry.  Detailed  exposition  of  such 
branches  as  shades  and  shadows,  perspective,  stereographic 
projection,  axonometry,  the  solution  of  spherical  triangles, 
etc.,  will  not  be  found  here,  but  the  student  is  prepared  to 
take  up  any  of  these  subjects. 


THE   MACMILLAN   COMPANY 

Publishers  64-66  Fifth  Avenue  New  Tork 


College  Algebra 


BY   E.    B.   SKINNER 

Associate  Professor  of  Mathematics  in  the  University  of  Wisconsin 
Edited  by  Professor  E.  R.  HEDRICK 

Cloth,  i2mo,  Illustrated,  $1.50 

The  essentials  of  a  course  in  algebra  are  put  into  the 
smallest  possible  compass  consistent  with  clearness  of 
presentation  to  elementary  students.  Enough  material  in 
the  way  of  examples  is  provided  to  enable  the  student  to 
acquire  some  skill  in  manipulation.  The  practical  side  of 
the  work  is  emphasized  by  giving  much  space  to  the  detailed 
study  of  the  functions  that  are  most  important  in  science  and 
in  applied  mathematics.  The  student  is  encouraged  to 
think  of  his  algebra  in  a  concrete  way  by  the  giving  of 
geometrical  interpretations  as  far  as  possible.  The  way  is 
prepared  not  only  for  a  clear  comprehension  of  the  imme- 
diately practical  side  of  algebra,  but  also  for  a  natural  and 
easy  transition  to  the  Analytical  Geometry  and  the  Cal- 
culus. 


THE   MACMILLAN   COMPANY 

Publishers  64-66  Fifth  Avenue  New  Tork 


Elementary  Mathematical  Analysis 
BY   JOHN    WESLEY   YOUNG 

Professor  of  Mathematics  in  Dartmouth  College 

AND 

FRANK    MILLET   MORGAN 

Assistant  Professor  of  Mathematics  in  Dartmouth  College 

Edited  by  EARLE  RAYMOND  HEDRICK 
Professor  of  Mathematics  in  the  University  of  Missouri 

$2.60,  Cloth,  1 2mo,  54.2 pages 

A  textbook  for  the  freshman  year  in  colleges,  universities, 
and  technical  schools,  giving  a  unified  treatment  of  the 
essentials  of  trigonometry,  college  algebra,  and  analytic 
geometry,  and  introducing  the  student  to  the  fundamental 
conceptions  of  calculus. 

The  various  subjects  are  unified  by  the  great  centralizing 
theme  of  functionality,  so  that  each  subject,  without  losing 
its  fundamental  character,  is  shown  clearly  in  its  relation- 
ship to  the  others,  and  to  mathematics  as  a  whole. 

More  emphasis  is  placed  on  insight  and  understanding  of 
fundamental  conceptions  and  modes  of  thought;  less 
emphasis  on  algebraic  technique  and  facility  of  manipula- 
tion. Due  recognition  is  given  to  the  cultural  motive  for 
the  study  of  mathematics  and  to  the  disciplinary  value. 

The  text  presupposes  only  the  usual  entrance  require- 
ments in  elementary  algebra  and  plane  geometry. 


THE   MACMILLAN  COMPANY 

Publishers  64-66  Fifth  Avenue  New  York 


Mathematics  for  Students  of  Agriculture 
and  General  Science 

BY  ALFRED  MONROE  KENYON 

Professor  of  Mathematics 

AND 

W.   V.    LOVITT 

Assistant  Professor  of  Mathematics  at  Purdue  University 
Edited  by  Professor  E.  R.  HEDRICK 

Cloth,  i2mo,  $2.00 

This  book  is  designed  as  a  text  in  Freshman  mathematics 
for  students  specializing  in  agriculture,  chemistry  and 
physics  in  colleges  and  in  technical  schools.  The  selection 
of  topics  has  been  determined  by  the  definite  needs  of  these 
students.  Since  this  course  is  to  constitute  the  entire 
mathematical  equipment  of  some  students,  some  chapters 
have  been  inserted  which  have  seldom  been  available  to 
Freshmen,  for  example,  the  chapters  on  annuities,  averages 
and  correlation,  and  the  exposition  of  Mendel's  Law  in  the 
chapter  on  the  binomial  expansion.  Particular  attention 
has  been  given  to  the  illustrative  examples  and  figures,  and 
to  the  grading  of  the  problems  in  the  lists.  Exercises 
constitute  about  one-fifth  of  the  text  and  contain  a  wealth 
of  material. 

Four  place  mathematical  tables  printed  at  the  end  of  the 
text  have  been  selected  and  arranged  for  practical  .use 
and  are  adapted  to  the  requirements  of  the  examples  and 
exercises  in  the  book. 


THE  MACMILLAN  COMPANY 

Publishers  64-66  Fifth  Avenue  New  York 


A  First  Course  in  Higher  Algebra 

BY  HELEN    A.    MERRILL 

Professor  of  Mathematics  in  Wellesley  College 

AND 

CLARA   SMITH 

Associate  Professor  of  Mathematics  in  Wellesley  College 

Cloth,  ismo,  $1.50 

At  this  time  when  combination  courses  are  receiving  so 
much  attention,  a  book  which  introduces  notions  of  the 
calculus  early  should  be  valuable.  The  present  volume,  an 
elementary  text  in  higher  algebra,  is  based  entirely  on  the 
theory  of  limits,  which  is  treated  as  rigorously  as  an  elemen- 
tary text  allows. 

It  is  intended  for  use  in  required  Freshman  mathematics 
courses,  and  it  seems  especially  well  adapted  for  this 
purpose.  Its  most  admirable  features  are  : 

1.  It  gives  the  student  the  ability  to  use  the  derivative 
in  a  course  in  analytic  geometry,  enabling  him  to  derive 
equations  of  tangents,  polars,  etc.,  in  the  best  possible  way. 

2.  The  sense  of  familiarity  which   the  student  gains 
with  the  beginnings  of  the  calculus  makes  for  ease  and 
rapidity  in  that  study. 

3.  He  is   equipped   earlier   than  is   customary   to   use 
mathematics  in  science  courses. 


THE   MACMILLAN   COMPANY 

Publishers  64-66  Fifth  Avenue  New  York 


14  DAY  USE 

RETURN  TO  DESK  FROM  WHICH  BORROWED 

LOAN  DEPT. 

This  book  is  due  on  the  last  date  stamped  below,  or 

on  the  date  to  which  renewed. 
Renewed  books  are  subject  to  immediate  recall. 


290cr  60M.ro 

?*  VfTO  L'U 

OtC  J9  15811 

-          110r.t'62MH 

r?nT^o  i  o 

r\CLw  LJ  t-i-^ 

OPT  r?     IQfi? 

UUi     /        i    r  c.. 

mJZ 

^ 

> 
V 

REC'D  LD 

FEB    5'64-HAI 

LD  21A-50m-4,'60 
CA9562slO)476B 


General  Library 

University  of  California 

Berkeley 


Q.A53\ 

P35 


.  •  •  <• 
THE  UNIVERSITY  OF  CALIFORNIA  LIBRARY 


